The average velocity in an interval [a, b] is
[f(b) - f(a)] / (b - a)
The instantaneous velocity is x = a is
f'(a)
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)
f(6) = 2(6)2 + 3(6) = 72 + 18 = 90 |
f(3) = 2(3)2 + 3(3) = 18 + 9 = 27 |
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
At t = 3 s'(3) = 4(3) + 3 = 12 + 3 = 15 m/sec |
At t = 6 s'(6) = 4(6) + 3 = 24 + 3 = 27 m/sec |
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)
f(5) = 16(5)2 = 16(25) = 400 |
f(3) = 16(3)2 = 16(9) = 144 |
= (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
When x = 3 m'(3) = √3/2√3 = 1/2 |
When x = 27 m'(27) = √3/2√27 = √3/2(3√3) = 1/6 |
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM