Displacement, Velocity, and Acceleration

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Example – 01:

  • A particle is moving in such a way that is displacement’s’ at any time ‘t’ is given by s = 2t2 + 5t +20. Find the velocity and acceleration of the particle after 2 seconds.
  • Solution:

The displacement of the particle is given by

s = 2t2 + 5t +20  ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 4t + 5 ……………… (2)



Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 4 ……………… (3)

To find velocity after 2 seconds i.e. t = 2 s

Substituting in equation (2)



Velocity = v = 4(2) + 5 = 8 + 5 = 13 unit/s

To find acceleration after 2 seconds i.e. t = 2 s

Acceleration = a = 4 units/s2

Ans: The velocity and acceleration of the particle after 2 seconds are 13 units/s and 4 units/s2 respectively.

Example – 02:

  • The displacement ‘s’ of a particle at a time ‘t’ is given by s = 5 + 20t – 2t2. Find its acceleration when its velocity is zero.
  • Solution:

The displacement of a particle is given by



s = 5 + 20t – 2t2  ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 20 – 4t  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = – 4 ……………… (3)



Given velocity is zero

20 – 4t = 0

4t = 20

T = 5 s

To find acceleration after 5 seconds i.e. t = 5 s



Acceleration = a = – 4 units/s2

Ans: The acceleration of the particle after 5 seconds is – 4 units/s2

Example – 03:

  • A particle is moving in such a way that is displacement’s’ at any time ‘t’ is given by s = t3 – 4t2 – 5t. Find the velocity and acceleration of the particle after 2 seconds.
  • Solution:

The displacement of the particle is given by

s = t3 – 4t2 – 5t  ……………… (1)



Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t2 -8t -5  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 6t – 8   ……………… (3)

To find velocity after 2 seconds i.e. t = 2 s

Substituting in equation (2)



Velocity = v = 3(2)2 -8(2) -5  = 12 – 16 – 5  = – 9 unit/s

To find acceleration after 2 seconds i.e. t = 2 s

Acceleration = a = 6(2) – 8 = 12 – 8 = 4 units/s2

Ans: The velocity and acceleration of the particle after 2 seconds are – 9 units/s and 4 units/s2 respectively.



Example – 04:

  • A particle is moving in such a way that is displacement’s’ at any time ‘t’ is given by s = 2t3 – 5t2 +  4t – 3. Find the time when acceleration is 14 ft/s2. Also, find velocity and displacement at that time.
  • Solution:

The displacement of the particle is given by

s = 2t3 – 5t2 +  4t – 3……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 6t2 – 10t + 4  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 12t – 10   ……………… (3)



Given acceleration = a = 14 ft/s2

Substituting in equation (3)

12t – 10 = 14

12t = 24

T = 2 s

Substituting t = 2 in equation (2)

Velocity = v = 6t2 – 10t + 4  = 6(2)2 – 10(2) + 4 =  24 – 20 + 4 = 8 ft/s

Substituting t = 2 in equation (1)

Displacement = s = 2(2)3 – 5(2)2 +  4(2) – 3 = 16 – 20 + 8 – 3 = 1 ft

Ans: After 4 s the acceleration will  be 14 ft/s2

The velocity is 8 ft/s and displacement is 1 ft

Example – 05:

  • A particle moves according to law s = t3 – 6t2 + 9t + 15, find the velocity when t = 0
  • Solution:

The displacement of the particle is given by

s = t3 – 6t2 + 9t + 15 ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t2 – 12t + 9  ……………… (2)

To find velocity at t = 0

Substituting t = 0 in equation (2)

Velocity = v = 3t(0)2 – 12t(0) + 9  = 9 units/s

Ans: The velocity at t = 0 is 9 units/s

Example – 06:

  • A particle moves under the law s = t3 – 4t2 – 5t. Find the displacement and velocity of particle when its acceleration is 4 units.
  • Solution:

The displacement of the particle is given by

s = t3 – 4t2 – 5t ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t2 – 8t – 5  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 6t – 8   ……………… (3)

Given acceleration = a = 4 units

Substituting in equation (3)

6t – 8 = 4

6t = 12

t = 2 s

Substituting t = 2 in equation (1)

Displacement = s = (2)3 – 4(2)2 – 5(2) = 8 – 16 – 10 = – 18 units

Substituting t = 2 in equation (2)

Velocity = v = 3(2)2 – 8(2) – 5 = 12 – 16 – 5 = – 9 units/s

Ans: When acceleration is 4 units displacement is -18 units and velocity is – 8 units/s

Example – 07:

  • A particle moves under the law s = t3/3 – t2/2 – t/2 +6. Find (i) its velocity at end of 4 s and (ii) acceleration and displacement when its velocity is 3/2 units.
  • Solution:

The displacement of the particle is given by

s = t3/3 – t2/2 – t/2 +6……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t2/3 – 2t/2 – 1/2 = t2 – t – 1/2      ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 2t – 1   ……………… (3)

(i) To find its velocity at end of 4 s, t = 4 s

Substituting t = 4 in equation (2)

Velocity = v = (4)2 – (4) – 1/2  = 16 – 4 – 1/2 = 11.5 units/s

(ii) To find acceleration and velocity when its velocity is 3/2 units.

t2 – t – 1/2 = 3/2

t2 – t – 2 = 0

(t – 2)(t + 1) = 0

t – 2 = 0 and t + 1 = 0

t = 2  and t = – 1

Time cannot be negative hence t = -1 not possible

t = 2 s

Substituting t = 2 in equation (3)

Acceleration =  2(2) – 1= 3 units/s2

Substituting t = 2 in equation (1)

Displacement = s = (2)3/3 – (2)2/2 – (2)/2 +6 = 8/3 – 2 – 1 + 6 = 17/3 units

Ans: Velocity at end of 4 s is 11.5 units/s

When velocity is 3/2 units, acceleration is 3 units/sand displacement is 17/3 units

Example – 08:

  • The displacement x of a particle at time t is given by x = 160 t – 16t2, show that its velocity at t = 1 and t =9 are equal in magnitude and opposite in direction.
  • Solution:

The displacement of the particle is given by

s = 160 t – 16t2……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 160 – 32t      ……………… (2)

Velocity at t = 1

Velocity = 160 – 32(1) = 128 units/s

Velocity at t = 9

Velocity = 160 – 32(9) = – 128 units/s

We can seet hat the velocities at t = 1 and t =9 are equal in magnitude and opposite in direction. (Proved)

Example – 09:

  • A particle is moving in a straight line and its displacement x from a fixed point O on the line at time t is given by . Show that the acceleration at time t is x-3.
  • Solution:

The displacement of the particle is given by

Acceleration

Thus acceleration at time t is x-3 (Proved as required)

Science > Mathematics > Applications of DerivativesYou are Here
Physics Chemistry  Biology  Mathematics

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