The maker of an automobile advertises that it takes 8 seconds to accelerate from 15 kilometers per hour to 70 kilometers per hour. Assuming constant acceleration, compute the distance, in meters, the car travels during the 8 seconds. Round your answer to two decimal places.
A) 49.50 m
B) 188.89 m
C) 94.44 m
D) 97.78 m
E) 61.11 m

Question

The maker of an automobile advertises that it takes 8 seconds to accelerate from 15 kilometers per hour to 70 kilometers per hour. Assuming constant acceleration, compute the distance, in meters, the car travels during the 8 seconds. Round your answer to two decimal places. ​
A) 49.50 m
B) 188.89 m
C) 94.44 m
D) 97.78 m
E) 61.11 m

Quizplus · Accepted Answer

To solve this problem, we can use the formula for distance under constant acceleration: $d = v_i t + \frac{1}{2} a t^2$, where $d$ is the distance, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. However, since we're not directly given the acceleration, we first need to find it using the formula $a = \frac{\Delta v}{t}$, where $\Delta v$ is the change in velocity.First, convert the velocities from km/h to m/s: $15 \, \text{km/h} = 15 \times \frac{1000}{3600} = 4.17 \, \text{m/s}$ and $70 \, \text{km/h} = 70 \times \frac{1000}{3600} = 19.44 \, \text{m/s}$.The change in velocity, $\Delta v = 19.44 \, \text{m/s} - 4.17 \, \text{m/s} = 15.27 \, \text{m/s}$.The acceleration, $a = \frac{\Delta v}{t} = \frac{15.27 \, \text{m/s}}{8 \, \text{s}} = 1.91 \, \text{m/s}^2$.Now, we can calculate the distance using the initial velocity and acceleration:$d = v_i t + \frac{1}{2} a t^2 = 4.17 \times 8 + \frac{1}{2} \times 1.91 \times 8^2 = 33.36 + 61.08 = 94.44 \, \text{m}$.Therefore, the car travels 94.44 meters during the 8 seconds.

The Maker of an Automobile Advertises That It Takes 8