A driver of a car traveling at 16.6 m/s applies
the brakes, causing a uniform deceleration of
3.0 m/s^2.
How long does it take the car to accelerate
to a final speed of 12.1 m/s?
Answer in units of s
part 2- How far has the car moved during the braking period?
Answer in units of m
Vf = Vi + a t
12.1 - 16.6 = -3 t
t = (16.6-12.1)/3
To find the time it takes for the car to accelerate to a final speed of 12.1 m/s, we can use the equation:
v = u + at
Where:
v = final velocity (12.1 m/s)
u = initial velocity (16.6 m/s)
a = acceleration (-3.0 m/s^2) [negative because it's deceleration]
t = time
First, let's rearrange the equation to solve for time (t):
t = (v - u) / a
Substituting the given values:
t = (12.1 m/s - 16.6 m/s) / (-3.0 m/s^2)
Now, let's calculate the time:
t = (-4.5 m/s) / (-3.0 m/s^2)
t = 1.5 s
Therefore, it takes the car 1.5 seconds to accelerate to a final speed of 12.1 m/s.
For the second part of the question, to determine how far the car has moved during the braking period, we can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement (distance)
u = initial velocity (16.6 m/s)
a = acceleration (-3.0 m/s^2) [negative because it's deceleration]
t = time (1.5 s)
Substituting the given values:
s = (16.6 m/s)(1.5 s) + (1/2)(-3.0 m/s^2)(1.5 s)^2
Now, let's calculate the distance:
s = 24.9 m + (-3.375 m)
s = 21.525 m
Therefore, the car has moved 21.525 meters during the braking period.