Question

Two particles move in a straight line together from rest, find the velocity at time = `t`?

Two particles `A` and `B` start to move at the same instant from a point `O`. The particles move in the same direction along the same straight line. The acceleration of `A` at time t `s` after starting to move is `a` `ms^-2`,
where `a = 0.05 − 0.0002t`.

Find `A`’s velocity when `t = 200` and when `t = 500`.

Answer 1

I got
`v_"200" = 2m*s^-1`
and,
`v_"500" = -25m*s^-1`

Explanation:

My problem here is that I think the mark scheme is wrong. I'm busy preparing for my AS Levels and this is from a Mechanics 1 Paper from 2015. The mark scheme answer to this is the following.

`v_"t" = 0.05t - 0.0001t^2`

then,

`v_"200" = 6m*s^-1`

and,

`v_"500" = 0m*s^-1`

Notice how in the equation they use 0.0001 instead of 0.0002. Did I miss something here or is the mark scheme wrong?

Answer 2

You are trying to find the velocity from the acceleration, where acceleration is the derivative of velocity. Therefore, the velocity can be found by integrating `a`. The integration looks like this:

`v(t)=int(0.05-0.0002t)dt`

`=>v(t)=0.05t-(0.0002t^2*1/2)`

I believe that forgetting the `1/2` is your problem!

This gives you:

`v(t)=0.05t-0.0001t^2`

Then you can put in your values for `t` :)