Kinematics

Question

A particle P moves in a straight line so that its displacement, s m, from a fixed point O , t seconds after passing through point A one the line , is given by s=t^3-3t^2-9t+30. Calculate

(a) the speed P when t=2,
(b) the acceleration of P when t=4
(c) the distance of P from A when it is instantaneously at rest,
(d) the average speed over the first 4 seconds.

Answer 1

s(t) = displacement function

v(t) = velocity/speed function

a(t) = acceleration function

 

In calculus, v(t) = s'(t), and a(t) = v'(t) = s''(t)

 

a)

s(t) = t^3 - 3t^2 - 9t +30

s'(t) = 3t^2 - 6t - 9

s'(2) = 3(2)^2 - 6(2) - 9 = -9

 

Since speed is a scalar quantity, the speed is the absolute vale of the velocity which is 9. 

 

b) 

s''(t) = 6t - 6

s''(4) = 6*4 - 6 = 18

 

c)

s'(t) = 0

3t^2 - 6t - 9 = 0

t^2 - 2t - 3 = 0

(t-3)(t+1) = 0

t=3

t=-1

At t=3, the particle rests. Its displacement at this time would be:

 

s(3) = 3^3 - 3(3)^2 - 9(3) + 30 = 3

s(0) = 0 - 0 - 0 + 30 = 30

 

From t=0 to t=3, the particle traveled a distance of 30-3 = 27

 

d) Average speed = s/t = [s(4) - s(0)]/[4 - 0]

s(4) = 4^3 - 3(4)^2 - 9(4) + 30 = 10

s(0) = 30

 

Average speed = (10-30)/(4-0) = -20/4 = -5

 

Since speed is a scalar quantity, the average speed would be 5.