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Relationship of Velocity, Acceleration and Displacement

The relationship of velocity, acceleration and displacement is obtained from equations of motion in a straight line.

Let us consider a body, which is moving in a straight line.

Let u = Initial velocity of the body in m/s

v = Final velocity of the body in m/s

t = Time in seconds, during which velocity changes from u to v

s = Distance travelled by the body in time t

a = Acceleration of the body.

5.4.1. Equation for Final Velocity

Change of velocity   = Final velocity – Initial velocity

= (v- u)

Relationship of Velocity, Acceleration and Displacement

5.4.4. Distance Travelled in the nth Second

Let  u = Initial velocity of a body

a = Acceleration of the body

Sn = Distance covered in n seconds

Sn-1 = Distance covered in (n – 1) seconds

Then distance travelled in the nth seconds

= Distance travelled in n seconds – Distance travelled in (n – 1) seconds

=Sn-Sn-1

Distance travelled in It seconds is obtained by substituting t = n in equation

s = ut + ½  at2 .

Problem 5.7.A body is moving with uniform acceleration and covers 15 m in fifth second and 25 m in 10th second. Determine:

(i)                           the initial velocity of the body, and

(ii)                        acceleration of the body. 

Sol. Given:

Distance covered in 5th second = 15 m

Distance covered in 10th second = 25 m

Let u = Initial velocity, and

a = Acceleration of the body.

Using equation (5.8) for the distance covered in nth second

5.4.5. Equation of Motions Due to Gravity. The acceleration due to gravity is ‘g’. Hence when a body falls, the equation of motions given by equations (5.5), (5.6) and (5.7) are modified by substituting ‘g’ in place of ‘a’. But when the body is moving vertically up the acceleration due to gravity is acting in the opposite direction. In that case the equations are modified by substituting (- g) in place of a. The value of g is taken as 981 cm/s2 or 9.81 m/s2. The distance ‘S’ is replaced by ‘h’. Hence the equations of motions due to gravity in the downward directions and upward directions becomes as :

5.4.6. Points to Be Remembered

i.            If a body starts from rest, its initial velocity is zero, i.e., u = 0.
ii.            If a body comes to rest, its final velocity is zero, i.e., v = 0.
iii.            If a body is projected vertically upwards, the final velocity of the body at the highest point is zero, i.e., v = 0.
iv.            If a body starts moving vertically downwards, its initial velocity is zero, i.e., u = 0.
v.            Acceleration due to gravity is taken positive when a body is moving vertically downwards. But if the body is moving vertically upwards, the acceleration due to gravity is taken negative.

Problem 5.8. A tower is 90 m in height. A particle is dropped from; the top of the tower and at the same time another particle is projected upward from the foot of the tower. Both the particle meet at a height of 30 m. Find the velocity, with which the second particle is projected upward. 

Fig. 5.2

Sol. Given:

Height of tower, h = 90 m.

Height at which both particle meet,

h1 = 30 m.

Distance travelled by first particle,

S1 =h-h1=90-30=60 m.

Distance travelled by the second particle,

S2 = h1 = 30 m.

For the first particle, we have (particle is moving down).

Initial velocity, u = 0

Distance, S1 = 60 m

Acceleration due to gravity, g = 9.81 m/s2.

Let the two particles meet after an interval of ‘t’ seconds.

Using the equation,

 

For the second particle, we have (particle is moving up)

Time, t=3.497s   (the time is same)

Distance, S2 = 30 m

Acceleration, g = – 9.81 m/s2

Let u = Initial velocity with which the second particle should be projected up.

Using the equation,

Problem 5.9.An aeroplane is flying horizontally at a height of800 m. A bomb is released from the aeroplane when the speed of aeroplane is 600 kmph (kilometre per hour). Determine the time required for the bomb to reach the ground and the horizontal distance travelled by the bomb during flight.

Sol. Given:

Height of aeroplane, h = 800 m

Speed of aeroplane, U = 600 kmph = 600 × 1000 m per hour

As the aeroplane is moving in the horizontal direction, when the bomb is released. This means the velocity of bomb, initially will be in the horizontal direction only. This velocity is equal to the velocity of aeroplane, i.e., 166.67 m/s, But at the same time, the bomb is attracted downwards due to gravitational effect. The initial velocity of the bomb in the downward direction is zero.

1st Case

Consider the motion of bomb in the downward direction.

Initial velocity in the downward direction,  u = 0

Height of bomb, when it is released, h = 800 m

Acceleration due to gravity,  g = 9.81 m/s2.

Let t = Time taken by bomb to reach the ground.

Time required to reach the ground=2.77 s.   Ans.

2nd Case

Horizontal distance travelled = Horizontal velocity of bomb × Time

 

 

 

Problem 5.10. A stone dropped into a well is heard to strike the water after 4 seconds. Find the depth of the well, if the velocity of sound is 350 m/sec.

(AMIE Summer, 1976)

Sol. Given:

Initial velocity of stone, u = 0

Time taken by stone to reach the bottom of well and time taken by the sound to reach from bottom of the well to the top of the well = 4 s.

Velocity of sound = 350 m/s.

Let h = Depth of well

t1 = Time taken by the stone to reach the bottom of well

t2 = Time taken by the sound to reach from bottom of the well to the top of the well.

Then   t1 + t2= 4 s.                                                                 …(i)

Consider the motion of the stone

Consider the motion of sound

When the stone strikes the water in the well the sound, produced by the stone, travels with uniform velocity of 350 m/s in all directions. Hence the time taken by the sound to reach the top of the well

 

Problem 5.11. A stone is thrown vertically upwards with a velocity of 19.6 m/ s from the top of a tower 24.5 m high. Calculate:

     i.            Time required for the stone to reach the ground.
    ii.            Velocity of the stone in its downward travel at the point in the same level as the point of projection.
iii.            The maximum height to which the stone will rise in its flight.

(AMIE Summer, 1977)

Sol. (i) Given:

Initial velocity of stone, u = 19.6 m/s

Height of tower, h = 24.5 m

Let AB is a tower and stone is thrown upwards from A with a velocity 19.6 m/s. The stone will reach the maximum height and after that will start moving downwards. Let the highest point to which the stone reaches is C as shown in Fig. 5.3. At
the point C the velocity of the stone will be zero.

Fig. 5.3

The total time taken by the stone to reach the ground is equal to the time taken by the stone to reach from A to C and then from C to D.

Let us first calculate the maximum height (i.e., AC) to which the stone will rise.

Let AC = h1 = Maximum height to which stone will rise.

For the motion of stone in the upward direction

u = 19.6 m/s, V = 0, Distance AC =h1

g = – 9.80 m/s2

Now using the equation

Maximum height to which the stone will rise, = 19.6 m. Ans.

(ii) The velocity of the stone when it is moving downward from C, at a point (E) which is in, the same level as A is equal to the velocity with which the stone is thrown upward from A, i.e.,

= 19.6 m/s. Ans. 

(iii) Time required for the stone to reach the ground

Let t1 = Time for the stone to move from A to C

t2 = Time for the stone to move from C to D

t = Total time for the stone to reach the ground

Then   t = t1 + t2                                                                                        … (i)

When the stone moves from A to C, we have

u = 19.6 m/s, v = 0

AC = h1 = 19.6 m, g = – 9.8 m/s2.

Time   t = t1.

Using the equation, v = u + gt, we have

0= 19.6 + (- 9.8) × t1 = 19.6 – 9.8t1

When the stone moves from C to D, we have

u = 0,             g = 9.8 m/s2

CD = Height BA + Height AC

= 24.5 + 19.6 = 44.1 m

Time, t = t2·

Problem 5.12. A stone is dropped from a height. After falling 5 seconds from rest, the
stone breaks the glass pane and in breaking, the stone loses
20% of its velocity. Find the distance travelled by the stone in the next second. Take g = 9.81 m/s2.

Sol. Given:

Initial velocity, u = 0

Time taken by stone in striking the glass pane = 5 s.

Acceleration due to gravity, g = 9.81 m/s2.

Let v = the velocity with which stone strikes the pane.

Using v = u + gt, we get

v = 0 + 9.81 × 5 = 49.05 m/s.

Velocity lost in breaking the stone

= 20% of the velocity with which the stone strikes

The velocity of the stone after breaking the glass pane

= 49.05 – 9.81 = 39.24 m/s.

Now the distance travelled by the stone, after breaking the glass pane, is one second is given by

(t = 1 see and initial velocity of stone after breaking = 39.24 m/s)

= 29.24 + 4.905 = 44.145 m. Ans.

Problem 5.13. An object falls from rest from an unknown height. In the last second of its motion, the object travels a distance of 53.90 m. If g = 9.80 m/s2, determine:

       i.            the height from which the object falls and
     ii.            total time taken by the object in falling. 

Sol. Given:

Initial velocity of the object, u = 0

Distance travelled in the last second = 53.90

Acceleration due to gravity, g = 9.80 m/s2.

Let h = Height from which the object falls and

t = Total time of fall.

Then h is also equal to the distance travelled by the object time ‘t’

Problem 5.14. Find the height of a tower from the top of which an object falls freely and during the last second of its motion, the object travels a distance equal to 2/3 of the height of the tower. Taking g = 9.80 m/s2.

Sol. Given:

Distance travelled in the last one second = (2/3) of the height of tower

Acceleration due to gravity, g = 9.80 m/s2

Initial velocity of the object, u = 0

Let h = Height of the tower

t = Total time of fall

But t is greater than unity, according to the data given in the problem.

t = 2.366 s.

Substituting the value of t in equation (i), we get

h = 4.9 × t2 = 4.9 × 2.3662 = 27.429 m

Height of the tower = h = 27.429 m . Ans. 

Problem 5.15. An object is thrown vertically upward, with a velocity of 30 m/s, Four
second later, a second object is project vertically upward with a velocity of
40 m/s. Determine:

i.            the time (after the first object is thrown) when the two objects will meet each other in air.
ii.            the height from the earth at which the two objects will meet.

Sol. Given:

Initial velocity of 1st object; u1 = 30 m/s.

Initial velocity of 2nd object, u2 = 40 m/s.

Let t = Time (after the first object is thrown) when the two objects will meet

h = Height at which the two objects will meet.

The height ‘h’ is covered by the first object in ‘t’ seconds whereas the second object covers the same height in (t – 4) seconds.

For the first object 

= 40(t – 4) – 4.905(t – 4)2                                                   … (ii)

Equating equations (i) and (ii), we get

30t – 4.905t2 = 40(t – 4) – .4.905(t – 4)2

30t – 4.905t2 – 40(t – 4) + 4.905(t – 4)2 = 0

30t – 40(t – 4) + 4.905 [(t - 4)2 - t2)] = 0

30t – 40t + 160 + 4.905[(t2 + 16 - 8t - t2) = 0

-10t + 160 + 4.905[16 - 8t] = 0

-10t + 160 + 78.48 – 39.24t = 0

49.24t = 238.48

Substituting the value of ‘t’ in equation (i), we get

h = 30 × 4.843 – 4.905 × 4.8432 = 145.29 – 115.045

= 30.245 m. Ans. 

 

Problem 5.16. A particle is dropped from the top of a tower 100 m high. Another particle is projected upwards at the same time from the foot of the tower, and meets the first particle at a height of 30 m. Find the velocity with which the second particle is projected Upwards. Take g=9.8m/s2.

Sol. Given:

Height of tower = 100 m

Initial Velocity of 1st particle, u1 = 0

Height from the foot of the tower at which the two particles meet = 30 m.

Distance travelled vertically up by the second particle, S2 = 30 m.

Distance travelled vertically down by the 1st particle,

S1 = 100 – 30 = 70 m.

Let t = Time, when the two particles meet

u2 = Initial velocity of the second particle, projected vertically up.

Case of the 1st particle.

The particle is dropped from the top of the tower.

Problem 5.17. A body is dropped from an unknown height. The ratio of the distances

covered by the falling body in the last and last but one second of its fall is 4/3 . If g = 9.8 m / S2, determine .

i.            the height from which the body is dropped and
ii.            the velocity with which the body will reach the ground.

Sol. (i) Given:

Initial velocity of the body, u = 0

Ratio of distances covered in last and last but one second of the fall of the body =4/3.

Acceleration due to gravity, g = 9.80 m/s2.

Let h = Height from which the body is dropped

t = Time of the fall of the body

h = Distance covered in (t – 1) second

h2 = Distance covered in (t – 2) second.

Then distance covered in last second = h – h1 and distance covered in the last but
one second = h1 – h2·

Now the distance covered in ‘t’ second is given by

 

(ii) Let v = Velocity with which the body will reach the ground.

v = u + gt, we get

v = 0 + 9.80 × 4.5                                       (u = 0)

= 44.1 m/s. Ans. 

Problem 5.18. A body falling freely under the action of gravity passes two points 20 m apart vertically in 0.4 seconds. From what height, above the higher point, did the body start to fall? Take g = 9.8 m/s2.

Sol. Given;

Initial velocity,  u = 0

Let the body starts from A and passes the two points B and C, 20 m apart in 0.4 seconds as shown in Fig. 5.4.

Let distance AB = x m

Then distance AC = (x + 20) m

If t = Time taken by the body to travel from A to B.

Then (t + 0.4) :: Time taken by the body to travel from A to C.

The distance ‘x’ travelled by the body in time ‘t’ is given by

5.4.7. Velocity and Acceleration of a Body Moving in a Straight Line by Differentiation. Let the equation of motion of a body moving in a straight line is given in terms of
displacement (s) and time (t) as s = t3 + 2t2 + 5t + 6                                     …(i)

By differentiating the above equation with respect to time, we get

= 3t2 + 4t + 5                                                                                                                    … (ii)

If the velocity at start is to be calculated, the value of t = 0 is to be substituted in equation (ii).

By differentiating the equation (ii) with respect to time, we get acceleration, i.e.,

The acceleration at start will be obtained by substituting t = 0, in equation (iii).

Problem 5.19. A particle moves along a straight line so that its displacement in metre from a fixed point is given by, s = t3 + 3t2 + 4t + 5.

Find:

  1.         i.            Velocity at start-and after 4 seconds
  2.      ii.            Acceleration at start and after 4 seconds.

Sol. Given:

S = t3 + 3t2 + 4t + 5

= 3t2 + 6t + 4                                                                                              … (i)

(i) Velocity at start will be obtained, if t = 0 is substituted in equation (i),

v(at t = 0) = 3 × 02 + 6 × 0 + 4 = 0 + 0 + 4

=4m/s. Ans.

Velocity after four seconds. Substitute t = 4 in equation.(i)
v(at t = 4) = 3 × 42 + 6 × 4+ 4

= 48 + 24 + 4 = 76 m/s. Ans.

(ii) Differentiating equation (i) with respect to time ‘t’, we get acceleration.

 

= 6t + 6.                                                                                                              …(ii)

Acceleration at start will be obtained, if t = 0 is substituted in equation (ii).

a(at t :: 0)      = 6 × 0 + 6 = 0 + 6

= 6 m/s2. Ans.

Acceleration after four seconds. Substitute t = 4 in equation (ii),

a(at t = 4)     = 6 × 4 + 6 = 24 + 6

= 30 m/s2 Ans.

Problem 5.20. The equation of motion of a particle moving in a straight line is given by

s = 18t + 3t2 – 2t3

where s is the total distance covered from the starting point is metres at the end of t seconds,

i.            the velocity and acceleration at start,
ii.            the time, when the particle reaches its maximum velocity and
iii.            the maximum velocity of the particle.

(AMIE Winter, 1982)

Sol. Given;

s = 18t + 3t2 – 2t3.

Differentiating both sides with respect to t,

Velocity, v = 18 + 6t – 6t2                                                              …(i)

(i) At start, t = 0. Hence velocity at start will be obtained by substituting t = 0 in equation (i),

Velocity at start,

v = 18 + 6 × 0 – 6 × 02 = 18 + 0 – 0

= 18 m/s. Ans.

Differentiating both sides of equation (i) with respect to t,

At start, t = 0. Hence acceleration at start will be obtained by substituting t = 0 in equation (iii).

Acceleration at start = 6 – 12 × 0 = 6 m/s2. Ans.

(ii) The velocity will be maximum, when the differentiation of the velocity with respect to time is equal to zero or dv/dt= 0. But dv/dt is given by equation (ii). Hence equating the equation (ii) to zero for maximum velocity, we get

6 – 12t = 0       or              12t = 6

(iii) From equation (i), velocity is given as

v = 18 + 6t – 6t2

For maximum velocity, t = ½ .

Hence substituting t = ½ in the above equation, we get maximum velocity as

= 21 – 1.5

= 19.5 m/s. Ans.

5.4.8. Velocity and Displacement of a Body Moving in a Straight Line by Integration. Let the equation of motion of a body moving in a straight line is given in terms of
acceleration (a) and time (t) as

a = t3 + 2t2 + 4t + 5                                           … (i)

 

If it is required to find the velocity and displacement at any time ‘t’, then equation (iii) and (v) are used.

Problem 5.21. A - particle moves along a straight line with an acceleration described by the equation a = – 8s-2 where a is in m/ s2 and sin m. When t = 1 s, s = 4 m and v = 2 m/ s. Determine acceleration when t = 2 s.

(Osmania, 1989)

Sol. Given:

Acceleration,            a = – 8s-2

When t = 1 s, s = 4 m and v = 2 m/s

Find acceleration when t = 2 s.

The acceleration in terms of velocity and displacement is given by equation (5.3),

Problem 5.22. A particle moves along a straight line with an acceleration described by the equation a = (4t2 -2), where a is in m/s2 and t is in seconds. What t = 0, the particle is located 2 m to the left of origin and when t = 2 seconds, the particle is 20 m to left of origin: Determine the position of the particle, when t = 4 seconds.

(AMIE Summer, 1988)

Sol. Given:

Acceleration,  a = 4t2 – 2

(If left of origin is considered – ve)

When t = 0, distance s = – 2 m

When t = 2 seconds, distance s = – 20 m

Find distance s when t = 4 seconds

Acceleration in terms of distance is given by,

Problem 5.23. A particle moves along a straight line with a velocity given by the equation

v = 2t3 - t2 – 2t + 4

where v is the velocity in m / s and t is time is seconds. When t = 2 seconds, the particle is found to be at a distance of to m from a station A. Determine:

(i)                           the acceleration and

(ii)                        displacement of the particle. after 6 seconds:

(Osmania, 1998)

Sol. Given:

Velocity, v = 2t3- t2- 2t + 4

When t = 2 seconds, distance s = 10 m

Find: (i) acceleration and (ii) displacement when t = 6 seconds

(i) Acceleration when t = 6 seconds

Problem 5.24. A body moves along a straight line and its acceleration a which varies
with t is given by
a = 2 – 3t. After 5 seconds from start of observations its velocity is observed to be 20 m/ s. After 10 seconds from start of observations the body was 85 metres from the, origin.

a)     Determine its acceleration, velocity and distance from the origin at the start of observations.

b)    Determine the time after start of observation in which the velocity becomes zero and its distance from the origin.

c)     Describe the motion diagrammatically.

(AMIE Summer, 1975)

Sol. Given :

Acceleration,  a = 2 – 3t                                                  … (i)

At  t = 5 seconds,  v = 20 m/s

At        t = 10 seconds,  s  = 85 m.

(a) (i) Acceleration at start will be obtained by substituting t = 0, in equation (i) as,
a = 2 – 3 × 0 = 2 – 0 = 2 m/s2. Ans.

(ii) The equation (i) can be written as

(c) Description of motion diagrammatically

The velocities for different times in sec is obtained from equation (iii). They are given as