On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.17 m/s2. (a) How long does such a car, initially traveling at 22.4 m/s, take to stop? (b) How far does it travel in this time?
time = (delta V)/(acceleration)
= 22.4/4.17 = 5.57 s
distance = time * (avg velocity)
= 5.57*11.2 = 60.2 m
To solve this problem, we can use the equations of motion. The first equation, known as the SUVAT equation, is:
v = u + at
where:
v = final velocity (0 m/s because the car stops)
u = initial velocity (22.4 m/s)
a = acceleration (deceleration in this case, -4.17 m/s^2)
t = time taken to stop (unknown)
(a) To find the time taken to stop, we can rearrange the equation as follows:
0 = 22.4 + (-4.17)t
Simplifying gives:
-4.17t = -22.4
Dividing both sides by -4.17:
t ≈ 5.37 seconds
Therefore, the car takes approximately 5.37 seconds to stop.
(b) To find the distance traveled, we can use the equation:
s = ut + 0.5at^2
where:
s = distance traveled (unknown)
Substituting the known values:
s = (22.4)(5.37) + 0.5(-4.17)(5.37)^2
Simplifying gives:
s ≈ 60.35 meters
Therefore, the car travels approximately 60.35 meters before coming to a stop.
To determine the time it takes for the car to stop, we can use the equation of motion:
v = u + at,
where:
v = final velocity (0 m/s, since the car has stopped),
u = initial velocity (22.4 m/s),
a = acceleration (deceleration in this case, -4.17 m/s^2), and
t = time.
(a) To find the time taken for the car to stop:
0 = 22.4 + (-4.17)t,
Simplifying the equation:
-4.17t = -22.4,
Divide both sides by -4.17:
t ≈ -22.4 / -4.17,
Approximately:
t ≈ 5.36 seconds.
The car takes approximately 5.36 seconds to stop.
(b) To find the distance traveled by the car in this time:
We can use the formula:
s = ut + (1/2)at^2,
where:
s = distance traveled,
u = initial velocity,
a = acceleration, and
t = time.
Plugging in the values:
s = (22.4)(5.36) + (1/2)(-4.17)(5.36)^2,
Simplifying the equation:
s = 119.74 + (1/2)(-4.17)(28.6896),
Simplifying further:
s = 119.74 - (1/2)(4.17)(28.6896),
Calculating:
s = 119.74 - (1/2)(119.74),
s = 119.74 - 59.87,
s ≈ 59.87 meters.
Therefore, the car travels approximately 59.87 meters before coming to a stop.