Expert: Ahmed Salami - 2/7/2009

Question
A particle with acceleration=2t-3 passes a point O with
velocity=2m/s , t=0 . find time t & displacement from O
when:a) it first comes to rest.b)it next comes to rest.
c)find total distance covered in first 2 seconds . thank u
.

Answer
Hi,
If acceleration, a = 2t - 3, the velocity function would be the integral of that function while the displacement function would be the subsequent integration (i.e of the velocity function). So,
v = t^2 - 3t + c
at t = 0, v = 2
2 = 0^2 - 3(0) + c
c = 2
therefore,
v = t^2 - 3t + 2
to find the displacement function s, we integrate
s = t^3/3 - 3t^2/2 + 2t + k
at t = 0, s = 0
and so k = 0
s = t^3/3 - 3t^2/2 + 2t
Now, the particle comes to rest at points where v = 0 i.e
t^2 - 3t + 2 = 0
(t-1)(t-2) = 0
t = 1s or 2s
the displacements at this times are
s = t^3/3 - 3t^2/2 + 2t
 = 1^3/3 - 3(1^2)/2 + 2(1)
 = 5/6 m
and
s = t^3/3 - 3t^2/2 + 2t
 = 2^3/3 - 3(2^2)/2 + 2(2)
 = 2/3 m

The total distance covered in the first 2s is the area under the velocity function graph(the definite integral) from t = 0 to 2. The curve is below the horizontal axis between t = 1 and 2, so we split the definite integral into two i.e
A1 = [t^3/3 - 3t^2/2 + 2t] from 0 to 1  
A2 = [t^3/3 - 3t^2/2 + 2t] from 1 to 2
A1 = [1^3/3 - 3(1^2)/2 + 2(1)] - [0^3/3 - 3(0^2)/2 + 2(0)]
  = 5/6
A2 = [2^3/3 - 3(2^2)/2 + 2(2)] - [1^3/3 - 3(1^2)/2 + 2(1)]
  = 2/3 - 5/6
  = -1/6
The negative value merely indicates that the area is below the horizontal axis and so we ignore it when we are adding areas.
The total distance covered is therefore,
s = 5/6 + 1/6
 = 1m

Regards