Mechanics

For (1), not that acceleration, a=dv/dt, and v=ds/dt. So a=dv/dt=d/dt(ds/dt)=d²s/dt². So if i want to go from acceleration to displacement i will need to 'undo' the double differentiation, i.e integrate twice. This will lead to two constants of integration, hence why you are given two initial conditions.

for (2), you are given an expression for the displacement from A at time t. B is at time t=25s away from A, can you use the expression to find the displacement B is from A?
If you are given an expression for s in terms of t, then how do you find acceleration?

For the next part, you are given the (constant) acceleration of the runner, can you find an expression for the displacement of the runner? Noting that the runner is going in the opposite direction of the car. If you have this expression, do you know what to do next? Again, you might find a sketch useful

So for the first question will we be integrating the equation that’s been given?

exactly that. if you post working i'm happy to have a look at it for you.

I’m confused with the second bit

Okay, so you have integrated d²s/dt² once to get to ds/dt. Now you need to integrate again to get s. Then from there you can put in the initial conditions to solve for the constants

Got this far

Okay, so first you started with acceleration which is d²s/dt². When you integrate once you go to v, which is ds/dt, with a constant of integration C. Which you got correctly as v=6t²-4t+C.

To get displacement, s, you need to integrate again, but this will introduce another constant of integration that might be different from C. As such, you shouldn't call it C. Perhaps you could label them as C₁,C₂ to make clear they are not the same constant of integration.

From there you can substitute in your initial conditions, and you should be able to solve for C₁ and C₂ giving you a final expression for the displacement, s.

Okay did that, I’m confused with what expression to use

Also did this for the second question but now I’m stuck

I agree with the 500m for the displacement between A and B, technically to find the distance you need to find the absolute value of this number, but its already positive so in this specific case the displacement is equal to the distance between A and B, but important to note.

For showing the acceleration is constant, if you are given an expression for displacement s, then how would you find an expression for the acceleration, a?

Suvat? I’m not sure

Remember that acceleration a=d²s/dt². Does this help?

So differentiate twice? And after that?

So differentiate twice? And after that?

Yes, d²s/dt² means differentiate s twice with respect to t, however you have already differentiated once to find ds/dt=v. So you only need to differentiate once more. Remember you are trying to show that acceleration is constant for all t.

What do you get when you work out that second derivative?

Got this

Exactly! If you want me to check over anything for the next part I can