Post

Created by @nathanedwards
 at November 2nd 2023, 4:20:39 pm.

Question

A car is traveling at a constant velocity of 30 m/s when the driver suddenly applies the brakes, causing the car to decelerate at a constant rate of -5 m/s^2. Assume the initial position of the car is 0 m.

a) Determine the time it takes for the car to come to a complete stop.

b) Calculate the distance traveled by the car during the deceleration.

Answer

a) To determine the time it takes for the car to come to a complete stop, we can use the** equation of motion**:

v=u+atv = u + at

where:
vv is the final velocity (0 m/s, since the car comes to a stop),
uu is the initial velocity (30 m/s),
aa is the acceleration (-5 m/s^2),
and tt is the time we want to solve.

Rearranging the equation, we get:

t=vuat = \frac{{v-u}}{a}

Substituting the given values into the equation:

t=0305=6st = \frac{{0-30}}{-5} = 6s

Therefore, it takes the car 6 seconds to come to a complete stop.

b) To calculate the distance traveled by the car during the deceleration, we can use the equation of motion:

s=ut+12at2s = ut + \frac{1}{2}at^2

where:
ss is the displacement or distance traveled,
uu is the initial velocity (30 m/s),
aa is the acceleration (-5 m/s^2),
and tt is the time taken (6 s in this case).

Substituting the given values into the equation:

s=(30m/s)(6s)+12(5m/s2)(6s)2s = (30 \text{m/s}) \cdot (6 \text{s}) + \frac{1}{2} (-5 \text{m/s}^2) \cdot (6 \text{s})^2

Simplifying the equation:

s=180m90m=90ms = 180 \text{m} - 90 \text{m} = 90 \text{m}

Therefore, the car travels a distance of 90 meters during the deceleration.