On a dry road, a car with good tires may

On a dry road, a car with good tires may be able to brake with a constant deceleration of $4.92 m/s^2$ . (a) How long does such a car, initially traveling at 24.6 m/s, take to stop? (b) How far does it travel in this time?

Solution:

To solve this question, we will use the following equation:

$v=v_0+at$ where $v$ is final velocity, $v_0$ is initial velocity, $a$ is acceleration and $t$ is time.

a)We will now substitute the values given to us in the question. Note that $v$ is equal to 0 because the car comes to a complete stop and the acceleration is negative because it’s deceleration.

$0=24.6m/s+(-4.92m/s^2)t$

$t=\frac{24.6m/s}{-4.92m/s^2}$

$t=5.00 s$

Therefore, the time it took for the car to stop is 5 seconds.

b) To find the distance the car travels in this time, we will use the following equation:

$v^2=v_0^2+2ax$ where $x$ is the position or distance.

$x=-\frac{(24.6 m/s)^2}{2(-4.92 m/s^2)}$

$x=61.5 m$

Therefore, the total distance the car traveled while the brakes were applied was 61.5 m.

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 28.

Solution:

To solve this question, we will use the following equation:

v=v_0+at where v is final velocity, v_0 is initial velocity, a is acceleration and t is time.

a)We will now substitute the values given to us in the question. Note that v is equal to 0 because the car comes to a complete stop and the acceleration is negative because it’s deceleration.

0=24.6m/s+(-4.92m/s^2)t

t=\frac{24.6m/s}{-4.92m/s^2}

t=5.00 s