Question

10

The velocity, Vm/s, of a particle moving in a straight line, at time t seconds is given by
V=1+9/t^2
for 1 is less than or equal to t which is less than or equal to 3. When t=3, the particle is 6m from a fixed point O
on the line.




Find the acceleration when t=2
Find an expression in terms of t for its distance from the fixed point.
Find the distance travelled between t = 1 and 1 = 3.
[4]​

Answer 1

Acceleration when time is
`2 \ sec` =
`-(5)/(4) \ m/s^2`

Distance
`s=9-(9)/(t)`

Distance travelled between
`t=1` and
`t=3` is
`6\ m`

Explanation:

Velocity
`V=1+(9)/(t^2) \ \ \ when \ \ 1\leq t\leq 3`

Acceleration(a): Rate of change of velocity.

`a=(dV)/(dt) =1-(18)/(t^3)`

at
`t=2`

`a=1-(18)/(8) \\=1-(9)/(4) \\=-(5)/(4)`

`a=-(5)/(4) \ m/s^2`

Distance(s):

`V=(ds)/(dt)\\\\(ds)/(dt)=1+(9)/(t^2)\\\\Integrate\ both\ sides\\\\s=1-(9)/(t)+k\\\\ k\ is\ a\ constant\\\\Given\ s=6\ when\ t=3\\\\6=1-(9)/(3)+k\\\\ k=8\\\\Hence\ s=1-(9)/(t)+8\\\\ s=9-(9)/(t)`

Distance between
`t=1 \ and \ t=3`

`=s(3)-s(1)\\\\=(9-(9)/(3))-(9-9) \\\\=6 \ m`