Question

1. (6 points) a car accelerates from an initial velocity of 6.0 m/s to a final velocity of 21 m/s at an acceleration of 3.35 m/s². a. how far does it travel while accelerating? b. how long is the acceleration period?

Explanation:

Step1: Identify the known - values

Initial velocity $u = 6.0\ m/s$, final velocity $v = 21\ m/s$, acceleration $a=3.35\ m/s^{2}$.

Step2: Use the equation $v^{2}-u^{2}=2as$ to find the distance $s$

We can re - arrange the equation $v^{2}-u^{2}=2as$ for $s$. So, $s=\frac{v^{2}-u^{2}}{2a}$.
Substitute the values: $v = 21\ m/s$, $u = 6.0\ m/s$, and $a = 3.35\ m/s^{2}$ into the formula.
\[

$$\begin{align*} s&=\frac{(21)^{2}-(6.0)^{2}}{2\times3.35}\\ &=\frac{441 - 36}{6.7}\\ &=\frac{405}{6.7}\\ &\approx60.45\ m \end{align*}$$

\]

Step3: Use the equation $v = u+at$ to find the time $t$

We can re - arrange the equation $v = u + at$ for $t$. So, $t=\frac{v - u}{a}$.
Substitute the values: $v = 21\ m/s$, $u = 6.0\ m/s$, and $a = 3.35\ m/s^{2}$ into the formula.
\[

$$\begin{align*} t&=\frac{21 - 6}{3.35}\\ &=\frac{15}{3.35}\\ &\approx4.48\ s \end{align*}$$

\]

Answer:

a. The car travels approximately $60.45\ m$.
b. The acceleration period is approximately $4.48\ s$.