College Physics Problem 2.28

A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.
(a) What is its average acceleration?
(b) How far does it travel in that time?

SOLUTION:

Part (a). The average acceleration of the motorcycle can be solved using the equation \overline{a}=\frac{\Delta v}{\Delta t}. Substitute the given into the equation. That is,

\overline{a}=\frac{\Delta v}{\Delta t}

\overline{a}=\frac{26.8\:m/s-0\:m/s}{3.90\:s}

\overline{a}=6.872\:m/s^2

Part (b). The distance traveled is equal to the average velocity multiplied by the time of travel. That is,

\Delta x=v_{ave}t

\Delta x=\left(\frac{0\:m.s+26.8\:m/s}{2}\right)\left(3.90\:s\right)

\Delta x=52.26\:m

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