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A truck travels along a straight, level road at the speed of 99ft/sec. when the driver is forced to apply the breaks to avoid an accident. If the truck undergoes a constant deceleration of 5ft/sec2, how far does the truck travel before coming to a complete stop?

I know I should be using integral to solve the question but i'm not sure about even set it up to solve it? What is the answer for this question using calculus?

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    $\begingroup$ Hint: at any given time, the position $x(t)$ should be a 2nd degree polynomial of $t$. $\endgroup$ – chi Feb 7 '17 at 0:49
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Just use the standard formulea of constant acceleraction. \begin{eqnarray*} v=u+at \\ s=ut+\frac{1}{2}a t^2 \end{eqnarray*} $u=99$,$a=-5$ & $v=0$. Use the first equation to find the time taken & then use the second equation to calculate $s$ the distance travelled. Alternatively just use \begin{eqnarray*} v^2 =u^2+2as \end{eqnarray*} Either way the answer is $980.1$ feet.

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Alternative hint (without calculus): since the acceleration is constant, the speed decreases linearly so $v = v_0 - a\,t$ where $v_0=99$ and $a = 5\,$. The truck comes at a stop when $v=0$ which means after a time $t = 99/5\,$. Since the speed varies linearly, the distance the truck traveled in time $t$ while braking is the same distance it would have traveled in the same time $t$ at average speed $(99+0)/2\,$.

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  • $\begingroup$ oh! so should use the equation d = v(t) + (1/2)at^2 to plug it in? $\endgroup$ – Kimmy.J Feb 7 '17 at 0:58
  • $\begingroup$ and what would the v(t) be? $\endgroup$ – Kimmy.J Feb 7 '17 at 0:59
  • $\begingroup$ @Kimmy.J Close: $v(t) = v_0 - a t$ and $d = \int_{t=0}^{t_0} v(t) dt = v_0 t_0 - a t_0^2/2\,$. $\endgroup$ – dxiv Feb 7 '17 at 1:00
  • $\begingroup$ so v(t) = 396/5 $\endgroup$ – Kimmy.J Feb 7 '17 at 1:01
  • $\begingroup$ @Kimmy.J $v(t)$ is a function of $t$, not a constant number. $\endgroup$ – dxiv Feb 7 '17 at 1:02

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