Physics
posted by Long .
Consider a car starting at rest at the beginning of a 250 m street. At time t = 2.0 s it is 20 m along the street. At time t = 4.0 s it is 80 m along the street. Find its average velocity in the:
(a) First 2.0 s (b) First 4.0 s (c) Last 2.0 s
My work:
I have done 2 question a and b and got the answer 10m/s and 20m/s respectively. However, i do not know how to explain and solve the last question. It is great If someone can help me to understand the last question.

To find the velocity in the last two seconds, we need to first find the time it takes to arrive at the end of the street of 250 length, and assuming uniform acceleration.
The distance travelled is given by
Δx = vi(t)+(1/2)at²
where
vi=initial velocity (m/s)
a=acceleration (m/s²)
t=time (s)
Since the car started from rest, vi=0,
At t=2, Δx=20m
20=0+(1/2)a(2²)
solving gives a=10 m/s²
Check at 4 sec.
Δx=0+(1/2)10(4²)=80m ok
Now we can assume uniform acceleration.
Calculate time to reach 250 m
250=(1/2)(10)t²
=> t=√50
So if we calculate where the car is located when t=√50  2, we have the distance travelled during the last two seconds, and can hence calculate the average velocity.
for t=√502,
Distance=(1/2)10(√50 2)²
=128.58 m
Average velocity
=(250128.58)m /2 s
=60.71 m/s