An automobile driver puts on the brakes and decelerates from 35.0 m/s to zero in 11.0 s.What distance does the car travel?
A)193 m
B)97 m
C)241 m
D)290 m
E)145 m

Question

An automobile driver puts on the brakes and decelerates from 35.0 m/s to zero in 11.0 s.What distance does the car travel?
A)193 m
B)97 m
C)241 m
D)290 m
E)​145 m

Quizplus · Accepted Answer

The distance traveled by the car can be found using the formula for distance under uniform acceleration, $d = v_i t + \frac{1}{2} a t^2$, where $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Here, $v_i = 35.0 \, \text{m/s}$, $t = 11.0 \, \text{s}$, and the final velocity $v_f = 0 \, \text{m/s}$. First, find the acceleration: $a = \frac{v_f - v_i}{t} = \frac{0 - 35.0}{11.0} = -3.18 \, \text{m/s}^2$. Then, calculate the distance: $d = 35.0 \times 11.0 + \frac{1}{2} \times (-3.18) \times (11.0)^2 = 385 - 192 = 193 \, \text{m}$.

An Automobile Driver Puts on the Brakes and Decelerates from 35.0