A car accelerates at a constant rate of 3.42 m/s2 from rest at the beginning of a freeway entrance ramp. At the end of the ramp the car has a speed of 25.49 m/s. How long is the ramp?
2 answers
s=v²/2a=25.49²/2•3.42 =95 m
just use this equation and remember that initial velocity is zero because the car began at rest
(final velocity)^2 = (initial velocity)^2 + 2*(acceleration)*(displacement)
you may need to rearrange the equation to say that
((final veocity)^2-(initial velocity)^2))/(2*accerleration) = d
so it would look like this
((25.49m/s)^2-(0m/s)^2)/(2*3.42m/s^2)
if you plug and chug you should be able to get the answer.
(final velocity)^2 = (initial velocity)^2 + 2*(acceleration)*(displacement)
you may need to rearrange the equation to say that
((final veocity)^2-(initial velocity)^2))/(2*accerleration) = d
so it would look like this
((25.49m/s)^2-(0m/s)^2)/(2*3.42m/s^2)
if you plug and chug you should be able to get the answer.
1 answer