A Level Maths Question- ​A particle P moves along a straight line such that at time t

For (a), I expanded it to get -2t^2 + 3t + 14. And I then differentiated to get -4t+3. Do I just do -4t + 3 = 0 to calculate t or am I oversimplifying it? I really appreciate the help, thanks

Yep! The derivative of velocity is acceleration, so the time at which P stops accelerating is the same as dv/dt = 0.

So t just equals 3/4?

Then what do I do for part (b)? Thanks for the help

Yeah don't forget to add units.

Use your expanded brackets and integrate. To find the value for c find t when v=0 (i think)

edit: sorry to find c use the equation for s at t=0 --> s=0 also at t=0

I would do it ever-so-slightly differently, but they're essentially the same thing and I think we get the same answer. First, work out the time at which P changes direction (i.e. v = 0), then do a definite integral of v(t) with a lower bound of 0 and an upper bound of t when v = 0.

So I worked out that t = 7/2 when v = 0 (please correct me if I’m wrong). And I used integration to get -2/3t^3 + 3/2t^2 +14t + c. What am I meant to do now? And how do I find C? Thanks for all the help 😃

c would be 0 as distance from 0 when t = 0 is 0.

Now sub in t = 3.5 into your integrated equation.