Question

While passing a slower car on the highway, you accelerate uniformly from 17.4 m/s to 27.3 m/s in a time of 10.0 s. How far do you travel during this time? What is your acceleration magnitude?

Answer 1

While passing a slower car on the highway, you accelerate uniformly from $17.4\,\mathrm{m/s}$ to $27.3\,\mathrm{m/s}$ in a time of $10.0\,\mathrm{s}$. How far do you travel during this time? What is your acceleration magnitude?

Solution.

The conditions are

$$\begin{array}{l} v_0 = 17.4\,\mathrm{m/s} \\ v_r(t_r) = 27.3\,\mathrm{m/s} \\ t_r = 10.0\,\mathrm{s} \\ \end{array}$$

Laws of motion are

$$\begin{array}{l} s(t) = v_0 t + \frac{a t^2}{2} \\ v(t) = v_0 + a t \\ \end{array}$$

Where $v(t)$ is velocity of car, $a$ is acceleration of car, $s(t)$ is path which travel car during time $t$, $v_0$ is initial velocity.

From hence

$$\begin{array}{l} a = \frac{v_r(t_r) - v_0}{t_r} = \frac{(27.3 - 17.4)\,\mathrm{m/s}}{10\,\mathrm{s}} = 0.99\,\frac{\mathrm{m}}{\mathrm{s}^2} \\ s(t_r) = v_0 t_r + \frac{a t_r^2}{2} = 17.4\,\frac{\mathrm{m}}{\mathrm{s}} \cdot 10\,\mathrm{s} + 0.99\,\frac{\mathrm{m}}{\mathrm{s}^2} \cdot \frac{(10\,\mathrm{s})^2}{2} = 223.5\,\mathrm{m} \\ \end{array}$$

Answer

$$\begin{array}{l} a = 0.99\,\frac{\mathrm{m}}{\mathrm{s}^2} \\ s(t_r) = 223.5\,\mathrm{m} \\ \end{array}$$