(a) $\text{40}\text{.}\text{0 km/h}$

(b) 34.3 km/h, $\text{25\xba}\phantom{\rule{0.25em}{0ex}}\text{S of E}\text{.}$

(c) $\text{average speed}=\text{3.20 km/h,}\phantom{\rule{0.25em}{0ex}}\stackrel{-}{v}=0.$

(a) $6\text{.}\text{61}\times {\text{10}}^{\text{15}}\phantom{\rule{0.25em}{0ex}}\text{rev/s}$

(b) 0 m/s

(a) $\text{16}\text{.}\text{5 s}$

(b) $\text{13}\text{.}\text{5 s}$

(c) $-2\text{.}{\text{68 m/s}}^{2}$

(a) $\text{20}\text{.}\text{0 m}$

(b) $-1\text{.}\text{00 m/s}$

(c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at $2\text{.}{\text{00 m/s}}^{2}$, then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.

(a) $\text{28}\text{.}\text{0 m/s}$

(b) $\text{50}\text{.}\text{9 s}$

(c) 7.68 km to accelerate and 713 m to decelerate

(a) $-\text{80}\text{.}4\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$

(b) $9\text{.}\text{33}\times {\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{s}$

(a) $7\text{.}\text{7 m/s}$

(b) $-\text{15}\times {\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$. This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high!

(a) $\text{32}\text{.}{\text{6 m/s}}^{2}$

(b) $\text{162 m/s}$

(c) $v>{v}_{\text{max}}$, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at $\text{32}\text{.}{\text{6 m/s}}^{2}$ during the last few meters, but substantially less, and the final velocity would be less than 162 m/s.

(a) $v=\text{12}\text{.}\text{2 m/s}$; $a=4\text{.}{\text{07 m/s}}^{2}$

(b) $v=\text{11}\text{.}\text{2 m/s}$

(a) ${y}_{1}=6\text{.}\text{28 m}$; ${v}_{1}=\text{10}\text{.}\text{1 m/s}$

(b) ${y}_{2}=\text{10}\text{.}\text{1 m}$; ${v}_{2}=5\text{.}\text{20 m/s}$

(c) ${y}_{3}=11\text{.}\mathrm{5\; m}$; ${v}_{3}=0\text{.300 m/s}$

(d) ${y}_{4}=10\text{.4 m}$; ${v}_{4}=-4\text{.60 m/s}$

(a) $a=-9\text{.}{\text{80 m/s}}^{2}$; ${v}_{0}=\text{13}\text{.}\text{0 m/s}$; ${y}_{0}=\text{0 m}$

(b) $v=0\text{m/s}$. Unknown is distance $y$ to top of trajectory, where velocity is zero. Use equation ${v}^{2}={v}_{0}^{2}+2a\left(y-{y}_{0}\right)$ because it contains all known values except for $y$, so we can solve for $y$. Solving for $y$ gives

Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.

(c) $2\text{.}\text{65 s}$

$v=\frac{(\text{11.7}-6.95)\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m}}{(40\text{.}\text{0 \u2013 20}.0)\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{238 m/s}$

- Use tape to mark off two distances on the track — one for cart
*A*before the collision and one for the combined carts after the collision. Push cart*A*to give it an initial speed. Use a stopwatch to measure the time it takes for the cart(s) to cross the marked distances. The speeds are the distances divided by the times. - If the measurement errors are of the same magnitude, they will have a greater effect after the collision. The speed of the combined carts will be less than the initial speed of cart
*A*. As a result, these errors will be a greater percentage of the actual velocity value after the collision occurs. (Note: Other arguments could properly be made for ‘more error before the collision' and error that ‘equally affects both sets of measurement.')

The position vs. time graph should be represented with a positively sloped line whose slope steadily decreases to zero. The *y*-intercept of the graph may be any value. The line on the velocity vs. time graph should have a positive *y*-intercept and a negative slope. Because the final velocity of the book is zero, the line should finish on the *x*-axis.

The position vs. time graph should be represented with a negatively sloped line whose slope steadily decreases to zero. The *y*-intercept of the graph may be any value. The line on the velocity vs. time graph should have a negative *y*-intercept and a positive slope. Because the final velocity of the book is zero, the line should finish on the *x*-axis.]