Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.700. What is the speed of the automobile after 2.00 s have elapsed? Ignore the effects of air resistance.
ma=μmg
a=μg
v=v₀-at=v₀-μgt=...
To find the speed of the automobile after 2.00 seconds have elapsed, we need to consider the deceleration caused by the friction between the tires and the road.
First, let's calculate the deceleration (negative acceleration) caused by the friction. This can be done using the formula:
Acceleration due to friction = coefficient of kinetic friction * acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2. So, the acceleration due to friction is:
Acceleration due to friction = 0.700 * 9.8 m/s^2 = 6.86 m/s^2
Next, we can use the following kinematic equation to find the final velocity of the automobile:
Final velocity = Initial velocity + (acceleration * time)
Since the driver suddenly locks the wheels, we can assume the initial velocity is 15.9 m/s (as given in the question), and the time is 2.00 s.
Final velocity = 15.9 m/s + (-6.86 m/s^2 * 2.00 s)
Final velocity = 15.9 m/s - 13.72 m/s
Final velocity = 2.18 m/s
Therefore, the speed of the automobile after 2.00 seconds have elapsed is 2.18 m/s.