A 1000Kg car is traveling at 40 m/s.
a.) if the car stops in 20 seconds, what is the deceleration rate?
b.) what friction force will stop the car in 20 seconds?
acceleration = (final velocity - initial velocity)/change in time
a = (0 - 40) m/s /20 s
a = - 2 m/s^2
F = m a
F = 1000(-2) = -2000 Newtons
To find the deceleration rate, you can use the formula:
deceleration = (final velocity - initial velocity) / time
a.) To find the deceleration rate, substitute the given values into the formula:
deceleration = (0 m/s - 40 m/s) / 20 s
deceleration = -40 m/s / 20 s
deceleration = -2 m/s²
So, the deceleration rate is -2 m/s². The negative sign indicates that the car is decelerating (slowing down).
b.) To find the friction force required to stop the car in 20 seconds, you can use Newton's second law of motion:
friction force = mass * acceleration
First, we need to find the acceleration using the deceleration rate:
acceleration = -2 m/s²
Now, substitute the values into the formula:
friction force = 1000 kg * (-2 m/s²)
friction force = -2000 N
The negative sign indicates that the friction force is exerted in the opposite direction of the car's motion, which is required to stop it.
Therefore, the deceleration rate of the car is -2 m/s² and the friction force required to stop the car in 20 seconds is -2000 N.