A motorcycle starting from rest, speeds up with a constant acceleration of 2.6 m/s2. After it has traveled 12o m, it slows down with a constant acceleration of −1.5 m/s until it attains a velocity of 12 m/s. What is the distance travelled by the motorcycle at that point? Hint. Do not use decimals, leave square roots indicated.
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1 Answer

s=½at² where s is distance, a is acceleration, t=time.

s=120, a=2.6, so t²=2s/a=240/2.6.

At that time v, velocity=at=2.6√(240/2.6)=√(240×2.6)=√624 m/s.

The motorcycle now decelerates, so a=-1.5, v=u+at gives us the final velocity v for initial velocity u, so we have v=12, u=√624 and a=-1.5. We have the formula v²-u²=2as, so:

144-624=-3s, therefore s=(624-144)/3=480/3=160m.

While the motorcycle was decelerating it covered a distance of 160m and it had already travelled 120m before decelerating, so the total distance for the whole trip was 160+120=280m.

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