M1 Constants of integration

a(t) for a 1D motion of a point is given by a(t)=3t^2+5t-6 for t<20 and a(t)=2t^3-t+1 for t>20. Find s when t=32.

The question essentially wants s(t) then plug in t=32 to find s, I think. First step must be to find v when t=20, so let's integrate:

v=t^3+(5/2)t^2-6t for t<20 (constant=0)
Evaluate this when t=20: v=8880 ms^(-1).
Now v(t) for t>20. v(t)=(1/2)t^4-(1/2)t^2+t+C

Here's my first stumbling block: how do we find the value of C now?

how do we find the value of C now?

The constant of integration is the initial velocity.

Although note that this isn't necessarily true in general - it so happens that the velocity here is a polynomial, so when t=0, every term cancels out so that the arbitrary constant is the only term left.
If, for example, we had $v(t) = cos(t)+C$, we'd have $v(0) = 1+C$, and then the arbitrary constant wouldn't be the initial velocity, but (the initial velocity)-1.

The constant of integration is the initial velocity.

Although note that this isn't necessarily true in general - it so happens that the velocity here is a polynomial, so when t=0, every term cancels out so that the arbitrary constant is the only term left.
If, for example, we had $v(t) = cos(t)+C$, we'd have $v(0) = 1+C$, and then the arbitrary constant wouldn't be the initial velocity, but (the initial velocity)-1.

Could you check through to see if my working makes sense? (The post above)

I haven't checked your numbers, but it looks fine to me. On a vaguely related note, something to be aware of is that they might try and catch you out by asking about distance rather than displacement - if you integrate velocity you'll get displacement, which isn't helpful in a distance question.

How would I solve this question if they asked for distance instead of displacement?

Then you'd have to look at the velocity and make sure that whenever the velocity was negative, you continued to add on the distance (rather than subtract, as you would implicitly be doing with displacement). That is, $x(t) = \int_0^t |v(u)| du$.

What is $u$?

A dummy variable that I just use for integration; it could just as well be $\alpha$ or something. I would usually use $x$, except that x already has a meaning (in this example, it's distance) so I can't use it again without being a bit sloppier than I would like.

Isn't it $t$?

Are you telling me I can't just integrate the modulus of v(t) with respect to t and get my answer?

Urgh, I find that this is quite a hard-to-explain point (a fact about my incompetence, I know, rather than the fact itself).
When you integrate, you integrate with respect to some variable. In the process, that variable "disappears" if you substitute stuff in (that is, if you're doing a definite integral). For example: $\int x^2 dx = \frac{x^3}3 + C$, but $\int_0^a x^2 dx = \frac{a^3}3$ - and you see that the x has vanished. Hence, if you do a definite integral, you need to use a "dummy variable": $\int_0^t x^2 dx = \frac{t^3}3$, and that's how you get t in your answer. Make sense? It's too late at night for me to be coherent :P

That's all very well, in fact it makes perfect sense - except that in your previous post you wrote $du$, which seems to conflict with the fact that you're still writing $dx$ when integrating f(x) here.

Maybe it would be best if you could just solve the example in my OP for distance? No need to get any numbers, just say the process.

In which case I shall rewrite my previous post as $\int_0^a u^2 du$ - the name I use for the variable doesn't matter, as long as it's a) consistent, and b) not used outside the integral.
In the example, you need to do $\int_0^t |\frac{u^4}2-\frac{u^2}2+u+C| du$ - of course, we don't necessarily have an initial condition any more, because the initial condition was about displacement rather than distance. You would carry out this integral by working out where the integrand became negative, splitting up the range of integration at those points, and evaluating the integrals separately there. Then you can add them together at the end. (Sorry, I'm a bit too tired at the moment to do it properly…)

a(t) for a 1D motion of a point is given by a(t)=3t^2+5t-6 for t<20 and a(t)=2t^3-t+1 for t>20. Find s when t=32.

The question essentially wants s(t) then plug in t=32 to find s, I think. First step must be to find v when t=20, so let's integrate:

v=t^3+(5/2)t^2-6t for t<20 (constant=0)
Evaluate this when t=20: v=8880 ms^(-1).
Now v(t) for t>20. v(t)=(1/2)t^4-(1/2)t^2+t+C

Here's my first stumbling block: how do we find the value of C now?

Quick question, those a's highlighted, what do they stand for?

Acceleration.