An object is travelling in a straight line. Its distance, s meters, from a fixed point at time t seconds is given by the expression

Question

$$s=t^3−t^2−6t$$

a) Find ds/dt when t=3 and interpret this result.

b) Find d^2s/dt^2 when t=3 and interpret this result.

c) Find the time in seconds when the velocity is 2m/s

(d) Using the results from 5(c) find the distance travelled when the velocity is 2m/s

A) $$S=t^3-t^2-6t$$ $$dy/dx = 3t^2-2t-6$$ $$When t = 3 = ds/df = 27-6-6 = 15ms^{-1}$$

B)

$$d^2s/dt^2 = 6t -2$$ $$=6(3)-2$$ $$=16$$

If i'm honest i don't know how to progress across the other questions from here. Can some show me how to do it/answer it? Really don't understand it

Answer 1

When you differentiate S w.r.t time, this would give you velocity at any time, you derive ds/dt, which is essentially distance and time divided to give speed. Equate the equation with 2, since the velocity at that moment is 2.

You will get a quadratic in t, which can then be solved to get the time. Now that you have the time, you can simply substitute it in the equation to get the displacement by that time :)

All the best

Answer 2

We have $$ s=t^3-t^2-6t $$ then \begin{align} \quad\left.\frac{ds}{dt}\right|_{t=3}&=3t^2-2t-6\qquad;\qquad\text{plug in}\ t=3\\ &=3(3)^2-2(3)-6\\ &=\cdots\\ \\ \quad\left.\frac{d^2s}{dt^2}\right|_{t=3}&=6t-2\qquad;\qquad\text{plug in}\ t=3\\ &=6(3)-2\\ &=\cdots\\ \\ \frac{ds}{dt}&=2\\ 3t^2-2t-6&=2\\ 3t^2-2t-8&=0\\ (t-2)(3t+4)&=0\\ t=2\quad&;\quad t=-\frac43\\ \\ s&=t^3-t^2-6t\qquad;\qquad\text{plug in}\ t=2\\ &=(2)^3-(2)^2-6(2)\\ &=\cdots \end{align} Interpret it the results by yourself.