Question

In Problems $66$ through $68$you are given the equation used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation.

b. Draw the before-and-after pictorial representation.

c. Finish the solution of the problem.

$$\begin{gathered} \frac{1}{2}(1500\mathrm{kg})(5.0\mathrm{m}/\mathrm{s}{)}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(10\mathrm{m}) \\ =\frac{1}{2}(1500\mathrm{kg})v_{\mathrm{i}}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(0\mathrm{m}) \end{gathered}$$

Short answer

(a) A $1500kg$car moves up a $10m$high hill and reaches the top with a speed of $5.0m/s$. What initial speed must the car have had at the bottom of the hill?

(b)

(c)$\text{Initialspeed}=14.9m/s$

Step-by-step solution

Given information (part a)

The following equation is given

$$\begin{gathered} \frac{1}{2}(1500\mathrm{kg})(5.0\mathrm{m}/\mathrm{s}{)}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(10\mathrm{m}) \\ =\frac{1}{2}(1500\mathrm{kg})v_{\mathrm{i}}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(0\mathrm{m}) \end{gathered}$$

Explanation (part a)

A $1500kg$car moves up a $10m$high hill and reaches the top with a speed of $5.0m/s$. What initial speed must the car have had at the bottom of the hill?

Given information (part b)

The following equation is given

$$\begin{gathered} \frac{1}{2}(1500\mathrm{kg})(5.0\mathrm{m}/\mathrm{s}{)}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(10\mathrm{m}) \\ =\frac{1}{2}(1500\mathrm{kg})v_{\mathrm{i}}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(0\mathrm{m}) \end{gathered}$$

Explanation (part b)

The pictorial representation of the car before and after.

Given information (part c)

The following equation is given

$$\begin{gathered} \frac{1}{2}(1500\mathrm{kg})(5.0\mathrm{m}/\mathrm{s}{)}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(10\mathrm{m}) \\ =\frac{1}{2}(1500\mathrm{kg})v_{\mathrm{i}}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(0\mathrm{m}) \end{gathered}$$

Explanation (part c)

The equation can be solved as below:

$$\begin{gathered} \frac{1}{2}(1500\mathrm{kg})(5.0\mathrm{m}/\mathrm{s}{)}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(10\mathrm{m}) \\ =\frac{1}{2}(1500\mathrm{kg})v_{\mathrm{i}}^2+(1500\mathrm{kg}\left)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right(0\mathrm{m}) \\ \left(750\mathrm{kg}\right)v_i^2=\left(18750\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2\right)+\left(147000\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2\right) \\ \left(750\mathrm{kg}\right)v_i^2=165750\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {v}_i^2=\frac{165750\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2}{750\mathrm{kg}} \\ {v}_i^{}=\sqrt{221{\mathrm{m}}^2/{\mathrm{s}}^2} \\ {v}_i^{}=14.9{\mathrm{m}}^{}/{\mathrm{s}}^{} \end{gathered}$$