(a) 7 m

(b) 7 m

(c) $+\mathrm{7\; m}$

a. $8\text{ m}+2\text{ m}+3\text{ m}=\underset{¯}{13\text{ m}}$

b. $\underset{¯}{9\text{ m}}$

c. $\Delta x=11\text{ m}-2\text{ m}=\underset{¯}{9\text{ m}}$

(a) $3\text{.}\text{0}\times {\text{10}}^4\text{m/s}$

(b) 0 m/s

$2\times {\text{10}}^7\text{years}$

$\text{34}\text{.}\text{689 m/s}=\text{124}\text{.}\text{88 km/h}$

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

(b) 34.3 km/h, $\text{25º}\text{S of E}\text{.}$

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

384,000 km

(a) $6\text{.}\text{61}\times {\text{10}}^{\text{15}}\text{rev/s}$

(b) 0 m/s

$4\text{.}\text{29}{\text{m/s}}^2$

(a) $1\text{.}\text{43 s}$

(b) $-2\text{.}\text{50}{\text{m/s}}^2$

(a) $\text{10}\text{.}8\text{m/s}$

(b)

38.9 m/s (about 87 miles per hour)

(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.

$0\text{.}\text{799 m}$

(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) $51\text{.}4\text{m}$

(b) $\text{17}\text{.}\text{1 s}$

(a) $-\text{80}\text{.}4{\text{m/s}}^2$

(b) $9\text{.}\text{33}\times {\text{10}}^{-2}\text{s}$

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

(b) $-\text{15}\times {\text{10}}^2{\text{m/s}}^2$. 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.

104 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) $y_4=10\text{.4 m}$; $v_4=-4\text{.60 m/s}$

$v_0=4\text{.}\text{95 m/s}$

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

1.91 s

(a) 94.0 m

(b) 3.13 s

(a) -70.0 m/s (downward)

(b) 6.10 s

(a) $\text{19}\text{.}\text{6 m}$

(b) $\text{18}\text{.}\text{5 m}$

(a) 305 m

(b) 262 m, -29.2 m/s

(c) 8.91 s

(a) $\text{115 m/s}$

(b) $5\text{.}{\text{0 m/s}}^2$

(a) 6 m/s

(b) 12 m/s

(c) ${\text{3 m/s}}^2$

(d) 10 s

(a) Car A is traveling faster at the checkpoint because it must go past the speed of car B to reach the same distance.

(b) i. Yes, the equation is consistent with the answer because the speed of car A is only a constant away from the correct answer. ii. Yes, the equation makes sense because $V=2V_0$.

(c)