Question

1. a car traveling at 10 m/s starts to decelerate steadily. it comes to a complete stop in 20 seconds. what is its acceleration? 2. an airplane travels down a runway for 4.0 seconds with an acceleration of 9.0 m/s². what is its change in velocity during this time? 3. a child drops a ball from a bridge. the ball strikes the water under the bridge 2.0 seconds later. what is the velocity of the ball when it strikes the water? 4. a boy throws a rock straight up into the air. it reaches the highest point of its flight after 2.5 seconds. how fast was the rock going when it left the boy’s hand?

Explanation:

Step1: Recall acceleration formula

The formula for acceleration is $a=\frac{v - u}{t}$, where $v$ is final - velocity, $u$ is initial - velocity, and $t$ is time. For the car, $u = 10\ m/s$, $v = 0\ m/s$, and $t = 20\ s$.
$a=\frac{0 - 10}{20}$

Step2: Calculate car's acceleration

$a=\frac{- 10}{20}=-0.5\ m/s^{2}$

For the airplane, the formula for change in velocity is $\Delta v=a\times t$. Given $a = 9.0\ m/s^{2}$ and $t = 4.0\ s$.
$\Delta v=9.0\times4.0 = 36\ m/s$

For the ball dropped from the bridge, we use the formula $v = u+gt$. Since it is dropped, $u = 0\ m/s$ and $g = 9.8\ m/s^{2}$, $t = 2.0\ s$.
$v=0 + 9.8\times2.0=19.6\ m/s$

For the rock thrown up, at the highest - point the final velocity $v = 0\ m/s$. Using the formula $v = u - gt$ (negative because the acceleration due to gravity acts downwards), $g = 9.8\ m/s^{2}$, $t = 2.5\ s$.
$0 = u-9.8\times2.5$, so $u=9.8\times2.5 = 24.5\ m/s$

Answer:

1. The car's acceleration is $-0.5\ m/s^{2}$.
2. The airplane's change in velocity is $36\ m/s$.
3. The ball's velocity when it strikes the water is $19.6\ m/s$.
4. The rock's initial velocity when it left the boy's hand is $24.5\ m/s$.