Question 57

3a. The function given is positive over the given interval, has a single root at the origin and a repeated root at $x=T$. This narrows the choices down to C and D. Note that $v=0$ for all $t>T$ and $t<0$ which eliminates C.

3b. $v=at(t-T)^2=a(t^3-2Tt^2+T^2t)$

$acceleration=\frac{dv}{dt}=a(3t^2-4Tt+T^2)=a(3t-T)(t-T)$

3c. The acceleration is a quadratic and hence is symmetrical and has a minimum value when $t$ is the mean of its roots. Therefore let $t=\frac{2T}{3}$.

$acceleration=a(2T-T)(\frac{2T}{3}-T)=\frac{-aT^2}{3}$

3d. Look at the graph - this must be for $t=0$. When $t=0$, $acceleration=aT^2$.

3e. The object doesn't move after $t=T$. Integrate the velocity between $t=0$ and $t=T$ to find the area under the curve and hence the distance travelled. $\int_0^Ta(t^3-2Tt^2+T^2t)dt=a[\frac{t^4}{4}-\frac{2Tt^3}{3}+\frac{T^2t^2}{2}]_0^T=a(\frac{T^4}{4}-\frac{2T^4}{3}+\frac{T^4}{2})=\frac{aT^4}{12}$

3a. The function given is positive over the given interval, has a single root at the origin and a repeated root at $x=T$. This narrows the choices down to C and D. Note that $v=0$ for all $t>T$ and $t<0$ which eliminates C.

3b. $v=at(t-T)^2=a(t^3-2Tt^2+T^2t)$

$acceleration=\frac{dv}{dt}=a(3t^2-4Tt+T^2)=a(3t-T)(t-T)$

3c. The acceleration is a quadratic and hence is symmetrical and has a minimum value when $t$ is the mean of its roots. Therefore let $t=\frac{2T}{3}$.

$acceleration=a(2T-T)(\frac{2T}{3}-T)=\frac{-aT^2}{3}$

3d. Look at the graph - this must be for $t=0$. When $t=0$, $acceleration=aT^2$.

3e. The object doesn't move after $t=T$. Integrate the velocity between $t=0$ and $t=T$ to find the area under the curve and hence the distance travelled. $\int_0^Ta(t^3-2Tt^2+T^2t)dt=a[\frac{t^4}{4}-\frac{2Tt^3}{3}+\frac{T^2t^2}{2}]_0^T=a(\frac{T^4}{4}-\frac{2T^4}{3}+\frac{T^4}{2})=\frac{aT^4}{12}$