# a cheetah can maintain a constant acceleration of 10.0 m/s2 for a time interval of 1.30 seconds. if the cheetah starts from rest, how long will it take the cheetah to move at 25.0 mph?

### James

Guys, does anyone know the answer?

get a cheetah can maintain a constant acceleration of 10.0 m/s2 for a time interval of 1.30 seconds. if the cheetah starts from rest, how long will it take the cheetah to move at 25.0 mph? from EN Bilgi.

## Solved A cheetah can maintain a constant acceleration of

Answer to Solved A cheetah can maintain a constant acceleration of

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## 3.4 Motion with Constant Acceleration

v0=10m/s,a=2.00m/s2${v}_{0}=10\phantom{\rule{0ex}{0ex}}\text{m/s},a=2.00\phantom{\rule{0ex}{0ex}}\text{m/}{\text{s}}^{2}$, and x = 200 m.

We need to solve for t. The equation x=x0+v0t+12at2$x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}$ works best because the only unknown in the equation is the variable t, for which we need to solve. From this insight we see that when we input the knowns into the equation, we end up with a quadratic equation.

We need to rearrange the equation to solve for t, then substituting the knowns into the equation:

200m=0m+(10.0m/s)t+12(2.00m/s2)t2.$200\phantom{\rule{0ex}{0ex}}\text{m}=0\phantom{\rule{0ex}{0ex}}\text{m}+\left(10.0\phantom{\rule{0ex}{0ex}}\text{m/s}\right)t+\frac{1}{2}\left(2.00\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2}\right){t}^{2}.$

We then simplify the equation. The units of meters cancel because they are in each term. We can get the units of seconds to cancel by taking t = t s, where t is the magnitude of time and s is the unit. Doing so leaves

200=10t+t2.$200=10t+{t}^{2}.$ We then use the quadratic formula to solve for t,

t2+10t−200=0t=−b±√b2−4ac2a,$\begin{array}{c}{t}^{2}+10t-200=0\\ \\ \\ t=\frac{\text{\u2212}b\pm \sqrt{{b}^{2}-4ac}}{2a},\end{array}$ which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the event happened 20 s before the motion began. We can discard that solution. Thus,

t=10.0s.$t=10.0\phantom{\rule{0ex}{0ex}}\text{s}\text{.}$

Source : **courses.lumenlearning.com**

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Guys, does anyone know the answer?