PROBLEMS ON AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY

Problem 1 :

A particle moves along a straight line in such a way that after t seconds its distance from the origin is

s = 2t² + 3t meters

(i) Find the average velocity between t = 3 and t = 6 seconds.

(ii) Find the instantaneous velocities at t = 3 and t = 6 seconds.

Solution :

i)  Let a = 3 and b = 6

= f(b) - f(a) / (b - a)

= f(6) - f(3) / (6 - 3)

average velocity = (90 - 27) / 3

= 63/3

= 21 m/sec

So, the average velocity is 21 m/sec.

ii)  s = 2t² + 3t

s'(t) = 2(2t) + 3(1)

= 4t + 3

Problem 2 :

A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of

s = 16t²

in t seconds.

(i) How long does the camera fall before it hits the ground?

(ii) What is the average velocity with which the camera falls during the last 2 seconds?

(iii) What is the instantaneous velocity of the camera when it hits the ground?

Solution :

i) The camera falls when it is in the height of 400 m

16t² = 400

t² = 400/16

t² = 25

t = 5

ii) Last two seconds means at t = 3 and t = 5

= [f(5) - f(3)] / (5 - 3)

= (400 - 144) / 2

= 256/2

= 128 ft/sec

So, in last two seconds the velocity is 128 ft/sec.

iii) When it hits the ground the velocity will become 0.

s(t) = 16t²

v(t) = 32t

v(5) = 32(5)

= 160 ft/sec

Problem 3 :

If the volume of a cube of side length x is v = x³ . Find the rate of change of the volume with respect to x when x = 5 units.

Solution :

v = x³

v'(t) = 3x²

Rate of change of volume at x = 5

v'(5) = 3(5)²

= 75 units

Problem 4 :

If the mass m(x) (in kilograms) of a thin rod of length x (in meters) is given by, m(x) = √3x then what is the rate of change of mass with respect to the length when it is x = 3 and x = 27 meters

Solution :

m(x) = √3x 

m(x) = √3√x 

Rate of change means, we have to find the first derivative.

m'(x) = √3(1/2√x)

= √3/2√x