Kinematics help a level

you basically have to double that 20m because it takes 20m to come to rest and then 20m back to come back to its original velocity (but opposite direction) and then travels 25m which you worked out in part a. So overall 65m.
I can show you a diagram of the boat in desmos:
https://www.desmos.com/calculator/rregdwrlbe
(just slide t from 0-25s)

I’m still not 100%.
I want to add this to my notes so how do I know when to double distance as it has appeared in another question too.

I’m still not 100%.
I want to add this to my notes so how do I know when to double distance as it has appeared in another question too.

The boat carries on moving forwards then stops and goes back. So it goes 20m forward then back to where it started ie 20m again ten another 25m to its final position. Part (a) 'hid' some of the motion as you are only looking at the final position.

Its worth being clear that the "s" in the suvat equations refers to the (signed) displacement from the origin/initial position, not the distance travelled. The magnitude of the displacement (unsigned) equals the distance travelled only if the velocity does not change sign (direction of motion reversed). As in mosaurlodons graph, when the velocity changes sign you have to add the distance travelled in both directions, so 20 (outwards) + 45 (backwards). So really if the question asks for the distance travelled, alarm bells should start ringing and you have to check if the velocity = 0 at a point during the motion and split the calculations at that point (or equivalently exploit the symmetry in the displacement parabola, so 2*20 + 25).
Here the initial velocity is 4m/s and the acceleration is -0.4m/s^2 so as they have opposing signs (opposite direction), it should be easy to see velocity is zero when t=10. Before that, velocity is positive, after that velocity is always negative. Edited the previous graph to include the value of the velocity at dispacement
https://www.desmos.com/calculator/m7iroikiyz
and it is positive initially, zero when s=20 and then becomes negative. Its parabolic due to the v^2=u^2+2as suvat relationship between velocity and displacement.

But in simple terms for my notes can I write when v becomes 0 total distance = 2 x distance +| displacement|

Thanks for your replies everyone and in an exam will they “hide” this in exams?

It would depend on the time interval over which youre calculating the distance travelled. Here the max displacement s=20 (velocity 0) occurs at t=10. It will return back to the origin at t=20 (symmetry of the parabola), so at that point, the distance travelled is 2*20. As the time 25>2*10, you double the distance of 20 and add the distance travelled in the last 5 seconds.
If you wanted the distance travelled up to 15 s, it would be simpler conceptually to work out the distance travelled from t=0s to 10s then add the distance travelled in the interval from t=10s to 15s, though there are a few equivalent ways of doing it.
A velocity-time sketch so
v = u+at
helps, noting the time (or displacement) when v=0 which is the critical point for such questions.

Can you show the parabola ?
Is it the v^2=u^2+2as formula?

This should be basic suvat stuff so for displacement-time you have
s = 4t - 0.2t^2
https://www.desmos.com/calculator/mlztqy7n3w
The max point is (10,20) and as the displacement returns back to zero and goes negative, you can do the 2*20+25 calculation for the total displacement as the distance travelled from 0 up to 20 and back again to 0 is 2*20.

I know and I’ve got 9 weeks and 6 days until my exams
I didn’t focus on mechanics at all.
I need at least a B for my exams. I know it’s a stupid question but is it possible?
I get it now btw but how do you know which equation to use?

This should be basic suvat stuff so for displacement-time you have
s = 4t - 0.2t^2 = -0.2(t-10)^2 + 20
https://www.desmos.com/calculator/mlztqy7n3w
The max point is (10,20) and as the displacement returns back to zero and goes negative, you can do the 2*20+25 calculation for the total displacement as the distance travelled from 0 up to 20 and back again to 0 is 2*20.

Ok I’m getting it kinda. Please don’t be annoyed but I’ve got more questions.
Could I do this with a v-t graph and find area under curve?

It would depend on the time interval over which youre calculating the distance travelled. Here the max displacement s=20 (velocity 0) occurs at t=10. It will return back to the origin at t=20 (symmetry of the parabola), so at that point, the distance travelled is 2*20. As the time 25>2*10, you double the distance of 20 and add the distance travelled in the last 5 seconds.
If you wanted the distance travelled up to 15 s, it would be simpler conceptually to work out the distance travelled from t=0s to 10s then add the distance travelled in the interval from t=10s to 15s, though there are a few equivalent ways of doing it.
A velocity-time sketch so
v = u+at
helps, noting the time (or displacement) when v=0 which is the critical point for such questions.

Ok I’m getting it kinda. Please don’t be annoyed but I’ve got more questions.
Could I do this with a v-t graph and find area under curve?