Intergration question (need help) A car starts from rest and it's acceleration after t seconds is given by the expression below.
Acceleration = $k - \frac 1 6 t$
It reaches a velocity of $35$ m/s at the end of $7$ minutes.
Find the distance travelled in the first minute.
You should express your answer in metres and to the nearest integer.
 A: Hints:

*

*Let $a(t)$, $v(t)$ and $x(t)$ be the acceleration, velocity and position respectively. Then $v(t)=\int a(t)dt$ and $x(t)=\int v(t)dt$.

*The velocity is 35m/s at the end of the 7 minute, and $t$ is in seconds.

*The car starts from rest, so what is $v$ when $t=0$? Now you can find $k$ and the constant of integration.

*Now integrate again to find $x(t)$ and use that $x(0)=0$, since the starting point is fixed (so can be taken to be $0$), to find the constant of integration.

*Finally find the distance travelled in the first minute, remember time is in seconds.

A: First, find out $k$ from
$$v(7)=35\times 60=\int_0^7 (k-1/6 t)dt$$
Then, calculate the distance with
$$d(1)= \int_0^1 v(t)dt =\int_0^1 \left(\int_0^t (k-1/6s)ds\right)dt$$
A: We are given
$$a = k-\frac{1}{6}t$$
We know that $a = \dfrac{\textrm dv}{\textrm dt}$. Hence,
$$\dfrac{\textrm dv}{\textrm dt} = k - \frac{1}{6}t$$
$$\implies \textrm dv = \left(k-\frac{1}{6}t\right)\,\textrm dt$$
Integrating both sides gives
$$\int\textrm dv = \int\left(k-\frac{1}{6}t\right)\,\textrm dt$$
$$v = kt - \frac{t^2}{12}+ C$$
To find $C$, we put $v=0, t=0$ as the car starts from rest,
$$0=0-0+C\iff C=0$$
Now,
$$v = kt - \frac{t^2}{12}$$
Now, we put $v = 35$ and $t=420$ (as $7$ minutes = $420$ seconds),
$$35 = 420k - \frac{420^2}{12}$$
$$k=\frac{14735}{420}=\frac{421}{12}$$
Thus, the final equation is:
$$v = \frac{421}{12}-\frac{t^2}{12}$$
$$v = \frac{421- t^2}{12}$$
Now, our required distance is given by the definite integral
$$s = \int_{0}^{1} v\,\textrm dt$$
$$s = \int_{0}^{1} \left(\frac{421-t^2}{12}\right)\,\textrm dt$$
$$s = \left[\frac{1}{12}\cdot \left(421t - \frac{t^3}{3}\right)\right]_{0}^{1}$$
$$s = \frac{1}{12}\cdot \left(421-1\right)$$
$$s = \frac{420}{12} = 35 \textrm{ m}$$
Hope this is clear :)
