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?
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.
To find the length of the ramp, we can use the kinematic equation relating acceleration (a), initial velocity (u), final velocity (v), and displacement (s):
v^2 = u^2 + 2as
In this case, the car starts from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) is given as 3.42 m/s^2, and the final velocity (v) is 25.49 m/s.
Let's substitute the known values into the equation and solve for displacement (s):
25.49^2 = 0^2 + 2(3.42)(s)
651.5401 = 6.84s
Dividing both sides of the equation by 6.84, we find:
s = 95.26 meters
Therefore, the length of the ramp is approximately 95.26 meters.