A driver of a car traveling at 15.9 m/s applies
the brakes, causing a uniform deceleration of
1.7 m/s2.
How long does it take the car to accelerate
to a final speed of 13.0 m/s?
Answer in units of s.
b) How far has the car moved during the
braking period?
Answer in units of m.
a) Review and apply the definition of acceleration.
a = (Change in velocity)/(time interval)
b) distance = (average velocity)*time
To solve this problem, we can use the equations of motion that relate the initial velocity (v0), final velocity (v), acceleration (a), time (t), and displacement (s).
For the first question, we are given:
v0 = 15.9 m/s (initial velocity)
v = 13.0 m/s (final velocity)
a = -1.7 m/s^2 (deceleration)
We can use the equation:
v = v0 + at
Rearranging the equation to solve for time (t):
t = (v - v0) / a
Plugging in the given values:
t = (13.0 m/s - 15.9 m/s) / (-1.7 m/s^2)
Calculating the result:
t = (-2.9 m/s) / (-1.7 m/s^2)
t ≈ 1.71 s
Therefore, it takes approximately 1.71 seconds for the car to accelerate to a final speed of 13.0 m/s.
For the second question, we need to find the distance traveled during the braking period. We can use the equation:
s = v0t + (1/2)at^2
Since the car is decelerating, the initial velocity will be the final velocity before braking, and the acceleration will be negative.
Given:
v0 = 15.9 m/s (initial velocity)
t = 1.71 s (braking time)
a = -1.7 m/s^2 (deceleration)
We can substitute these values into the equation to calculate the distance traveled (s):
s = (15.9 m/s)(1.71 s) + (1/2)(-1.7 m/s^2)(1.71 s)^2
Calculating the result:
s = 27.489 m + (1/2)(-1.7 m/s^2)(2.9241 s^2)
s = 27.489 m + (-2.484 m)
s ≈ 25.005 m
Therefore, the car has moved approximately 25.005 meters during the braking period.