A cyclist starts at the top of a straight slope with an initial velocity of 2.82 m/s. Five seconds later, she is at the bottom of the incline, having traveled 45.8 m. Find her velocity at the bottom of the hill and her acceleration, assuming it to be constant. (Let down the hill be the positive direction. Indicate the direction with the sign of your answer.)

Question

A cyclist starts at the top of a straight slope with an initial velocity of 2.82 m/s. Five seconds later, she is at the bottom of the incline, having traveled 45.8 m. Find her velocity at the bottom of the hill and her acceleration, assuming it to be constant. (Let down the hill be the positive direction. Indicate the direction with the sign of your answer.)

Answer 1

To find the cyclist's velocity at the bottom of the hill and her acceleration, we can use the following kinematic equations:

v = u + at
s = ut + (1/2)at^2

where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement

Given:
u = 2.82 m/s (initial velocity)
t = 5 s (time)
s = 45.8 m (displacement)

First, let's find the acceleration using the second equation.

s = ut + (1/2)at^2

Plugging in the known values:

45.8 = 2.82(5) + (1/2)a(5)^2

To simplify, let's solve for a:

a = (2(45.8 - 2.82(5)) / (5)^2

a = (91.6 - 28.2) / 25

a = 63.4 / 25

a ≈ 2.54 m/s^2

Now, let's find the final velocity by plugging the values into the first equation:

v = u + at

v = 2.82 + (2.54)(5)

v = 2.82 + 12.7

v ≈ 15.52 m/s

Therefore, the cyclist's velocity at the bottom of the hill is approximately 15.52 m/s in the positive direction. Her acceleration is approximately 2.54 m/s^2 in the positive direction.