The maker of an automobile advertises that it takes 7 seconds to accelerate from 30 kilometers per hour to 70 kilometers per hour.Assuming constant acceleration,compute the distance,in meters,the car travels during the 7 seconds.Round your answer to two decimal places.
A)41.14 m
B)194.44 m
C)97.22 m
D)54.44 m
E)38.89 m

Question

The maker of an automobile advertises that it takes 7 seconds to accelerate from 30 kilometers per hour to 70 kilometers per hour.Assuming constant acceleration,compute the distance,in meters,the car travels during the 7 seconds.Round your answer to two decimal places. ​
A)41.14 m
B)194.44 m
C)97.22 m
D)54.44 m
E)38.89 m

Quizplus · Accepted Answer

To find the distance traveled during acceleration, we first convert the speeds from km/h to m/s: $30 \, \text{km/h} = \frac{30 \times 1000}{3600} = 8.33 \, \text{m/s}$ and $70 \, \text{km/h} = \frac{70 \times 1000}{3600} = 19.44 \, \text{m/s}$. The acceleration $a$ can be found using the formula $a = \frac{\Delta v}{\Delta t} = \frac{19.44 \, \text{m/s} - 8.33 \, \text{m/s}}{7 \, \text{s}} = 1.59 \, \text{m/s}^2$. The distance $s$ traveled under constant acceleration is given by $s = v_0t + \frac{1}{2}at^2$, where $v_0$ is the initial velocity. Substituting the values, we get $s = 8.33 \times 7 + \frac{1}{2} \times 1.59 \times 7^2 = 58.31 + 38.915 = 97.225 \, \text{m}$, which rounds to $97.22 \, \text{m}$.

The Maker of an Automobile Advertises That It Takes 7