Question

Can I get answers to this A level past question please

Answer 1

(a) Consider that the distance function s(t) can be obtained by integrating the function v(t), as follow:

`s(t)=\int v(t)dt`

under the condition that s(1) = 50 m.

(b) By integrating you have:

`\begin{gathered} s(t)=\int (6t^2-4t+10)dt \\ s(t)=6(t^3)/(3)-4(t^2)/(2)+10t+C \\ s(t)=2t^3-2t^2+10t+C \end{gathered}`

use the IVP s(1) = 50 and solve for C:

`\begin{gathered} s(1)=2(1)^3-2(1)^2+10(1)+C=50 \\ s(1)=2-2+10+C=50 \\ 10+C=50 \\ C=40 \end{gathered}`

Then, the function s(t) is:

`s(t)=2t^3-2t^2+10t+40`

(c) For t=10s you have:

`\begin{gathered} s(10)=2(10)^3-2(10)^2+10(10)+40 \\ s(10)=2000-200+100+40 \\ s(10)=1940 \end{gathered}`

The distance traveled in 10 s is 1940 m.

(d) The acceleration is the first derivative of v(t), then, you have:

`\begin{gathered} a(t)=v(t)^(\prime)=12t-4 \\ a(5)=12(5)-4=60-4=56 \end{gathered}`

The acceleration is 56m/s^2