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# The position of a car moving on a straight highway is given by$x=t^2-\frac{t^3}{90} m$where t is the time in seconds. Determine (a) the distance traveled by the car before it comes to a stop; and (b) the maximum velocity reached by the car.

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In this problem, we are asked to find the distance traveled by car before it stops and the maximum velocity reached by car. To solve this problem, the first thing to do is to differentiate the equation with respect to time $(t)$ to find the x. Since the velocity is a rate of change of displacement with time and one way of expressing that is to say $v=\dfrac{d_x}{d_t}.$ We can use this to differentiate the position of the car.

\begin{aligned} \dfrac{d_x}{d_t}&=\dfrac{d\Biggr(t^2-\dfrac{t^3}{90}\Biggr)}{d_t}\\\\ &=\dfrac{d(t^2)}{d_t}-\dfrac{d\Biggr(\dfrac{t^3}{90}\Biggr)}{d_t}\\\\ &=2t-\dfrac{3t^2}{90}&~~~~~~~~~~~~~~~~(1) \end{aligned}

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