A runner of mass 51.5 kg starts from rest and accelerates with a constant acceleration of 1.35 m/s2 until she reaches a velocity of 5.7 m/s. She then continues running with this constant velocity. (Take the direction the runner is going to be the positive direction. Indicate the direction with the sign of your answer.)

(a) How far has she run after 64.7 s?
(b) What is the velocity of the runner at this point?

do this in two parts:

distance running speeding up
vf=vi+2at solve for t, then
distance= vf/2 * t

second runnign at constant velocity
distance= vf*(64.7-t)

add the distances.

To solve these questions, we need to break the problem down into different stages and use the relevant equations of motion.

Stage 1: Acceleration phase
We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given:
Initial velocity, u = 0 m/s (starting from rest)
Acceleration, a = 1.35 m/s^2
Final velocity, v = 5.7 m/s

Using the equation v = u + at, we can rearrange it to solve for the time, t:
t = (v - u) / a

t = (5.7 m/s - 0 m/s) / 1.35 m/s^2
t = 4.22 s

So, the time taken to reach the final velocity is 4.22 seconds.

Stage 2: Constant velocity phase
After reaching the velocity of 5.7 m/s, the runner continues running with this constant velocity.

(a) How far has she run after 64.7 s?
In the constant velocity phase, the distance traveled can be found using the equation: s = ut + 0.5at^2, where s is the distance traveled, u is the initial velocity, and a is the acceleration (which is 0 during this phase).

Given:
Initial velocity, u = 5.7 m/s
Time, t = 64.7 s

Using the equation s = ut + 0.5at^2, we can calculate the distance traveled after 64.7 seconds:
s = (5.7 m/s)(64.7 s) + 0.5(0)(64.7 s)^2
s = 369.99 m

Therefore, the runner has run a distance of 369.99 meters after 64.7 seconds.

(b) What is the velocity of the runner at this point?
The runner is still moving at a constant velocity of 5.7 m/s during this phase. So, the velocity is 5.7 m/s.

To solve this problem, we can use the kinematic equations of motion. The two equations we need are:

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

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = distance

(a) To find how far the runner has run after 64.7 seconds, we need to calculate the distance traveled during the acceleration phase and the distance traveled during the constant velocity phase.

First, let's calculate the distance traveled during the acceleration phase. The runner starts from rest, so the initial velocity (u) is 0 m/s. The final velocity (v) is 5.7 m/s, and the acceleration (a) is 1.35 m/s^2. We can use the equation v = u + at to find the time (t) taken during the acceleration phase:

v = u + at
5.7 = 0 + 1.35t
t = 5.7 / 1.35
t ≈ 4.22 s

Now, let's calculate the distance (s1) during the acceleration phase using the equation s = ut + (1/2)at^2:

s1 = (0)(4.22) + (1/2)(1.35)(4.22)^2
s1 = 0 + (1/2)(1.35)(17.8084)
s1 ≈ 12.02 m

Next, let's calculate the distance traveled during the constant velocity phase. The velocity remains at 5.7 m/s for the remaining time of 64.7 - 4.22 = 60.48 seconds. We can use the equation s = ut + (1/2)at^2, where the acceleration is now 0 m/s^2:

s2 = (5.7)(60.48) + (1/2)(0)(60.48)^2
s2 = 344.736 + 0
s2 = 344.736 m

Therefore, the total distance traveled by the runner after 64.7 seconds is s = s1 + s2:
s = 12.02 + 344.736
s ≈ 356.756 m

So, the runner has run approximately 356.756 meters after 64.7 seconds.

(b) To find the velocity of the runner at this point, we can use the equation v = u + at with t = 64.7 s and a = 1.35 m/s^2:

v = u + at
v = 0 + (1.35)(64.7)
v ≈ 87.345 m/s

The positive sign indicates that the runner is moving in the positive direction.

Therefore, the velocity of the runner at this point is approximately 87.345 m/s.