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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 = 2t
^{2}+ 5t +20. Find the velocity and acceleration of the particle after 2 seconds. **Solution:**

The displacement of the particle is given by

s = 2t^{2} + 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/s^{2}

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

#### Example – 02:

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

The displacement of a particle is given by

s = 5 + 20t – 2t^{2} ……………… (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/s^{2}

**Ans:** The acceleration of the particle after 5 seconds is – 4 units/s^{2}

#### Example – 03:

- A particle is moving in such a way that is displacement’s’ at any time ‘t’ is given by s = t
^{3}– 4t^{2}– 5t. Find the velocity and acceleration of the particle after 2 seconds. **Solution:**

The displacement of the particle is given by

s = t^{3} – 4t^{2} – 5t ……………… (1)

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

Velocity = v = ds/dt = 3t^{2} -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/s^{2}

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

#### Example – 04:

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

The displacement of the particle is given by

s = 2t^{3} – 5t^{2} + 4t – 3……………… (1)

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

Velocity = v = ds/dt = 6t^{2} – 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/s^{2}

Substituting in equation (3)

12t – 10 = 14

12t = 24

T = 2 s

Substituting t = 2 in equation (2)

Velocity = v = 6t^{2} – 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/s^{2}

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

#### Example – 05:

- A particle moves according to law s = t
^{3}– 6t^{2}+ 9t + 15, find the velocity when t = 0 **Solution:**

The displacement of the particle is given by

s = t^{3} – 6t^{2} + 9t + 15 ……………… (1)

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

Velocity = v = ds/dt = 3t^{2} – 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 = t
^{3}– 4t^{2}– 5t. Find the displacement and velocity of particle when its acceleration is 4 units. **Solution:**

The displacement of the particle is given by

s = t^{3} – 4t^{2} – 5t ……………… (1)

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

Velocity = v = ds/dt = 3t^{2} – 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 = t
^{3}/3 – t^{2}/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 = t^{3}/3 – t^{2}/2 – t/2 +6……………… (1)

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

Velocity = v = ds/dt = 3t^{2}/3 – 2t/2 – 1/2 = t^{2} – 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.

t^{2} – t – 1/2 = 3/2

t^{2} – 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/s^{2}

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/s^{2 }and displacement is 17/3 units

#### Example – 08:

- The displacement x of a particle at time t is given by x = 160 t – 16t
^{2}, 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 – 16t^{2}……………… (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

Thus acceleration at time t is x^{-3} **(Proved as required)**

Science > Mathematics > Applications of Derivatives > You are Here |

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