The derivative can also be used to determine the rate of change of one variable with respect to another. A few examples are population growth rates, production rates, water flow rates, velocity, and acceleration.

A common use of rate of change is to describe the motion of an object moving in a straight line. In such problems, it is customary to use either a horizontal or a vertical line with a designated origin to represent the line of motion.

On such lines, movements in the forward direction considered to be in the positive direction and movements in the backward direction is considered to be in the negative direction.

**Example 1:**

A missile fired ground level rises x meters vertically upwards in t seconds and x = 100t - (25/2)t^{2}. Find

(i) the initial velocity of the missile,

(ii) the time when the height of the missile is a maximum

(iii) the maximum height reached and

(iv) the velocity with which the missile strikes the ground.

**Solution :**

x = 100t - (25/2)t²

(i) The time when the missile starts is 0.

The distance is changing with respect to time.

dx/dt = 100 (1) - (25/2) (2t)

= 100 - 25 t

= 100 - 25 (0)

= 100 - 0

dx/dt = 100 meter/seconds

(ii) When a object reaches its maximum height the velocity will become zero.

dx/dt = 0

dx/dt = 100 - 25 t

100 - 25 t = 0

- 25 t = - 100

t = 100/25

t = 4 seconds

So, the object is taking 4 seconds to reach the maximum height.

(iii) The missile is taking 4 seconds to reach its maximum height. By applying the value 4 for t, we get the distance covered by the missile.

put t = 4

x = 100t - (25/2)t²

x = 100 (4) - (25/2) (4)²

= 400 - (25/2) (16)

= 400 - (25) (8)

= 400 - 200

= 200 meter

(iv) When the missile reaches the ground, the height of the missile = 0

x = 100t - (25/2)t^{2}

100t - (25/2)t^{2} = 0

- (25/2)t² = - 100 t

(25/2)t² = 100 t

t²/t = 100 (2/25)

t = 200/25

t = 8

dx/dt = 100 - 25 t

= 100 - 25 (8)

= 100 - 200

= -100 meter/seconds

Since it reaches the ground the answer is having negative sign.

**Example 2 :**

A particle of unit mass moves so that displacement after t seconds is given by x = 3 cos (2t - 4). Find the acceleration and kinetic energy at the end of 2 seconds. (K.E = (1/2) m v²)

**Solution :**

displacement of a particle in seconds

x (t) = 3 cos (2t - 4)

mass = 1

to find acceleration we have to different the given equation two times

velocity dx/dt = 3 [- sin (2t - 4)] (2(1) - 0)

Velocity = -6 sin (2t - 4)

velocity at t = 2

v = -6 sin (2(2) - 4)

= -6 sin (4-4)

= -6 sin (0)

= 0

Acceleration d^{2}x/dt^{2} = -6 cos (2 t - 4) (2(1)- 0)

= -6 cos (2 t - 4)2

= -12 cos (2 t - 4)

now we have to put t = 2

= -12 cos (2(2) - 4)

= -12 cos (4 - 4)

= -12 cos (0)

= -12 (1) ==> -12

Kinetic energy K.E = (1/2) m v^{2}

= (1/2) (1) (0)^{2}

= (1/2) (0)

= 0

**Example 3 :**

The distance x meters traveled by a vehicle in time t seconds after the brakes are applied is given by

x = 20 t - (5/3)t^{2}

Determine

(i) the speed of the vehicle (in km/hr) at the instant the brakes are applied and

(ii) the distance the car traveled before it stops.

**Solution :**

the distance x meters traveled by a vehicle in time t seconds

x = 20 t - (5/3)t^{2}

To find the speed of the vehicle, differentiate it with respect to "t"

dx/dt = 20 (1) - (5/3)(2t)

= 20 - (10 t/3)

the speed of the vehicle (in km/hr) at the instant the brakes are applied

t = 0

= 20 - (10(0)/3)

= 20 - 0

to velocity after the brakes are applied

= 20 meter/seconds

to convert this into km/hr we have to multiply it with the fraction 3600/1000

= (20 ⋅ 3600)/1000

= 72 km/hr

We hope that the students would have understood how to solve application problems on rate of change.

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