A car accelerates from $5.0 \mathrm {~m} / \mathrm { s }$ to $21 \mathrm {~m} / \mathrm { s }$ at a constant rate of $3.0 \mathrm {~m} / \mathrm { s } ^ { 2 }$ How far does it travel while accelerating?
A)69 m
B)207 m
C)41 m
D)117 m

Question

Quizplus · Accepted Answer

The distance traveled during acceleration can be calculated using the formula $d = \frac{v^2 - u^2}{2a}$, where $v = 21 \mathrm{m/s}$ is the final velocity, $u = 5 \mathrm{m/s}$ is the initial velocity, and $a = 3.0 \mathrm{m/s^2}$ is the acceleration. Plugging in the values, we get $d = \frac{(21)^2 - (5)^2}{2 \times 3} = \frac{441 - 25}{6} = \frac{416}{6} = 69 \mathrm{m}$.

A Car Accelerates From 5.0 m/s5.0 \mathrm {~m} / \mathrm { s }5.0 m/s

A Car Accelerates From $5.0 \mathrm {~m} / \mathrm { s }$