Question

QUESTION 6

Answer 1

(a) The instant velocity and acceleration at a specific time $t_0$ are given by evaluating the first and second derivatives at $t=t_0$

So, we have

$\begin{cases}x(t)=2t^2+3t-5\\x'(t)=4t+3\\x''(t)=4\end{cases}$

Which implies

$\begin{cases}x'(5)=23\\x''(5)=4\end{cases}$

So, when time is 5 seconds, the object has a velocity of 23 m/s and an acceleration of 4 m/s^2 (which is indeed constant along all motion).

(b) The equation for the displacement is already given: $s=e^{3t}$. So, we simply have to evaluate this function at t=0 to get

$s(0)=e^{3\cdot 0}=e^0=1$

So, the displacement at t=0 is 1 unit along the positive side of the line.

As for the acceleration, applying the derivation formula

$\dfrac{d}{dx} e^{f(x)} = e^{f(x)}\cdot f'(x)$

we have

$s'(t) = 3e^{3t},\quad s''(t) = 3\cdot 3e^{3t}=9e^{3t}$

So, we have

$s(t) = e^{3t},\quad s''(t) =9e^{3t}$

and thus the acceleration is nine times the distance.

(c)

We have to evaluate the integral

$\displaystyle \int_0^T \sin(2x)$

The antiderivative of sin(2x) is -1/2cos(2x), becase we have

$\dfrac{d}{dx} -\dfrac{1}{2}\cos(2x) = \dfrac{1}{2}2\sin(2x)=\sin(2x)$

So, we have

$\displaystyle \int_0^T \sin(2x) = \left[-\dfrac{1}{2}\cos(2x)\right]_0^T = -\dfrac{1}{2}\cos(2T)+\dfrac{1}{2}\cos(0)=\dfrac{1-\cos(2T)}{2}$