Question

A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60

what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.

Answer 1

Approximately
`2\; \rm s`, assuming that the floor of this parking lot is level,
`\mu_(\rm k) = 0.60`, and
`g = 9.81\; \rm m\cdot s^(-2)`.

Step-by-step explanation:

Let
`m` denote the mass of this vehicle. Weight of this vehicle:
`m\, g`.

If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight:
`N = m \, g`.

Given that
`\mu_(\rm k)`, the kinetic friction between this vehicle and the ground would be consistently
`\mu_(\rm k) \, N = \mu_(\rm k) \, m \, g` until the vehicle comes to a stop.

Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be
`(-\mu_(\rm k) \, m \, g)` (negative because this force is opposite to the direction of the motion.)

By Newton's second law of motion, the acceleration of this vehicle would be:

`\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= (-\mu_(\rm k) \, m \, g)/(m) \\ &= -\mu_(\rm k)\, g \\ &= -0.60 * 9.81\; \rm m\cdot s^(-2) \\ &= -5.886\; \rm m\cdot s^(-2)\end{aligned}`.

In other words, braking would reduce the velocity of this vehicle by a constant
`5.886\; \rm m\cdot s^(-1)` every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from
`v_(0) = 12\; \rm m\cdot s^(-1)` to
`v_(1) = 0\; \rm m\cdot s^(-1)`:

`\begin{aligned}t &= (v_(1) - v_(0))/(a) \\ &= (0\; \rm m\cdot s^(-1) - 12\; \rm m\cdot s^(-1))/(-5.886\; \rm m \cdot s^(-2)) \\ &\approx 2.0\; \rm s \end{aligned}`.

Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.