A car moves in a straight line with constant acceleration. Starting from rest at t=0 it travels 8.0 m in 4.0 s. What is the speed of the car at t= 4 s.
PLEASE explain :)
d = 0.5a*t^2.
8 = 0.5a*4^2, a = 1 m/s^2.
V = Vo + a*t.= 0 + 1*4 = 4 m/s.
To find the speed of the car at t=4 s, we need to use the equations of motion related to constant acceleration.
There are three relevant equations of motion:
1. v = u + at
This equation relates the final velocity (v) to the initial velocity (u), acceleration (a), and time (t).
2. s = ut + (1/2)at^2
This equation relates the displacement (s) to the initial velocity (u), time (t), and acceleration (a).
3. v^2 = u^2 + 2as
This equation relates the final velocity (v) to the initial velocity (u), displacement (s), and acceleration (a).
Let's solve this problem using equation 2 since we have the initial velocity, time, and displacement. Rearranging the equation, we have:
s = ut + (1/2)at^2
Plug in the values we have:
8.0 m = 0 m/s * 4.0 s + (1/2) * a * (4.0 s)^2
Now we can solve for a (acceleration). Rearranging the equation, we get:
(1/2) * a * (4.0 s)^2 = 8.0 m
Multiply both sides by 2:
a * (4.0 s)^2 = 16.0 m
Divide by (4.0 s)^2:
a = 16.0 m / (4.0 s)^2 = 4.0 m/s^2
Now we know the value of acceleration.
To find the final velocity, we can plug the values into equation 1:
v = u + at
Plug in the values we have:
v = 0 m/s + (4.0 m/s^2)(4.0 s)
v = 0 m/s + 16.0 m/s
v = 16.0 m/s
Therefore, the speed of the car at t=4 s is 16.0 m/s.