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.
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

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.

by Top Rated User (763k points)

Related questions

0 answers
asked Feb 11, 2013 in Word Problem Answers by anonymous | 313 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
85,089 questions
90,220 answers
59,424 users