A boy sledding down a hill accelerates at 1.24
m/s2
.
If he started from rest, in what distance
would he reach a speed of 8.88 m/s?
Answer in units of m
1) You should have been taught about the equations relating speed acceleration, time and distance.
One is simply speed = acceleration * time, v =a*t.
Another is one that says distance covered is average velocity * time. since the sled starts at rest, and acceleration is constant, the average speed is half the final speed. Is that undersandable? average speed = 1/2 a*t.
and lastly the distance covered is the average speed times time, d = 1/2 a*t*t (see how we got that?).
now we don't know time, so the trick is to manipulate the equations so that we don't need time. We do have final speed and accel, and we're looking for distance.
first rearrange v+a*t into t = v/a. now we're going to plug that into d = 1/2 a *t^2 (t^2 = t squared = t*t) gives
d = 1/2 a * (v/a)^2. There you go, and eq for distance from a and v.
formula:
vf^2=vi^2+2as