An automobile is traveling 65 km/hr. It brakes decelerate it 6.o m/s^2.
a) How long will it take to stop the car?
b) How far will the car travel after the brakes are applied?
3.01s
10.83
To solve this problem, we can use the equations of motion.
a) To find the time it will take to stop the car, we need to determine the initial velocity (u), final velocity (v), and the deceleration (a). In this case, the initial velocity is 65 km/hr, which we need to convert to m/s.
We know that 1 km/hr = 1000 m/3600 s, so
65 km/hr = 65 * 1000 m / 3600 s = 18.06 m/s.
Therefore, the initial velocity (u) is 18.06 m/s, the final velocity (v) is 0 m/s (since the car comes to a stop), and the deceleration (a) is -6.0 m/s^2 (negative because it slows the car down).
We can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Rearranging the equation, we have t = (v - u) / a.
Plugging in the values, we get:
t = (0 - 18.06) / -6.0
t = 18.06 / 6.0
t = 3.01 seconds
Therefore, it will take approximately 3.01 seconds to stop the car.
b) To find the distance the car will travel after the brakes are applied, we can use the equation: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values, we get:
s = (18.06 * 3.01) + (1/2)(-6.0)(3.01)^2
s = 54.4406 - 27.099
s = 27.3416 m
Therefore, the car will travel approximately 27.34 meters after the brakes are applied.
To answer these questions, we need to use the kinematic equations. The first step is to convert the speed from km/hr to m/s.
a) To find the time it takes to stop the car, we can use the kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s, as the car comes to a stop)
u = initial velocity (65 km/h)
a = acceleration (-6.0 m/s^2, since the car is decelerating)
t = time
First, let's convert the initial velocity from km/h to m/s:
65 km/h * (1 hr/3600 s) = 18.06 m/s (rounded to 2 decimal places)
Plugging the values into the equation:
0 m/s = 18.06 m/s + (-6.0 m/s^2)t
Rearranging the equation to solve for time:
-18.06 m/s = -6.0 m/s^2 * t
t = -18.06 m/s / -6.0 m/s^2
t ≈ 3.01 seconds
So, it will take approximately 3.01 seconds to stop the car.
b) To find the distance traveled, we can use another kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity (18.06 m/s)
a = acceleration (-6.0 m/s^2)
t = time (3.01 seconds)
Plugging the values into the equation:
s = 18.06 m/s * 3.01 s + (1/2)(-6.0 m/s^2)(3.01 s)^2
s ≈ 54.26 meters (rounded to 2 decimal places)
Therefore, the car will travel approximately 54.26 meters after the brakes are applied.