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 = (1/2) a t^2
8 = .5 a (16)
so a = 1 m/s^2
v = a t = 1 (4) = 4 m/s
or use head like this
average speed = 8 m/4s = 2 m/s
begin speed is zero
so final speed = twice average speed = 2*2 = 4 m/s
To find the speed of the car at t=4 s, we can use the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the car starts from rest (u=0), the initial velocity is 0.
We know that the car travels a distance of 8.0 m at t=4.0 s. We can use this information to find the acceleration.
To find the acceleration, we can use another kinematic equation:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity
a = acceleration
t = time
Plugging in the values, we have:
8.0 = 0 + (1/2)a(4.0^2)
8.0 = 8a
Simplifying the equation, we find:
a = 1 m/s^2
Now, we can substitute the values of a and t into the first kinematic equation to find the final velocity:
v = 0 + (1)(4.0)
v = 4.0 m/s
Therefore, the speed of the car at t=4 s is 4.0 m/s.