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 car's initial velocity is $5.0 \mathrm {~m} / \mathrm { s }$, its final velocity is $21 \mathrm {~m} / \mathrm { s }$, and its acceleration is $3.0 \mathrm {~m} / \mathrm { s } ^ { 2 }$. We can use the following equation to calculate the distance traveled while accelerating:$$d = \frac{1}{2} (v_i + v_f) t$$where $d$ is the distance traveled, $v_i$ is the initial velocity, $v_f$ is the final velocity, and $t$ is the time taken to accelerate.We don't know the time taken to accelerate, but we can use the following equation to find it:$$v_f = v_i + at$$where $a$ is the acceleration.Substituting this into the first equation, we get:$$d = \frac{1}{2} (v_i + v_f) \left( \frac{v_f - v_i}{a} \right)$$Simplifying this, we get:$$d = \frac{(v_f^2 - v_i^2)}{2a}$$Substituting the given values, we get:$$d = \frac{(21 \mathrm {~m} / \mathrm { s })^2 - (5.0 \mathrm {~m} / \mathrm { s })^2}{2(3.0 \mathrm {~m} / \mathrm { s } ^ { 2 })}$$$$d = 69 \mathrm {~m}$$Therefore, the car travels 69 m while accelerating.

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

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