AP Physics 1- Ernst 1.2 pg 13, tt's 19,24,32,39,40,46,49,71,78 19. A sports car moving at a constant velocity travels 120m in 5.0s. If it then brakes and comes to a stop in 4.0s, what is the magnitude of its acceleration (assumed to be constant) inm/s^2 and in g's (g=9.80(m)/(s)) 24. A light plane must reach a speed of 35m/s for takeoff. How long a runway is needed if the (assume to be constant) acceleration is 3.0m/s^2 32. A driver is traveling 18.0m/s when she sees a red light ahead.Her car is capable of braking at a rate of 3.65m/s^2 If it takes her 0.350 s to get the brakes on and she is 20 Om when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in what direction (before the stop light, or into the intersection)? 39. A stone is dropped from the top of a cliff.It is seen to hit the ground below after 3.55 s. How high is the cliff? 40. Estimate: (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380m high)in the movie, and (b) his velocity just before "landing." 46. A stone is thrown vertically upward with speed of 24.0m/s (a) How fast is it moving when it is at a height of 13.0 m? (b) How much time is required to reach this height? (c) Why are there two answers to part b? 49. A helicopter is ascending vertically with a speed of 5.40m/s At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? (Hint: what is v_(i) for the package?) 71. A stone is thrown vertically upward with a speed of 15.5m/s from the edge of a cliff 75.0m high. (a) How much later does it reach the bottom of the cliff? (b) What is its speed before hitting? (c) What total distance did it travel? 78. The graph shows the position vs. time graph for two bicycles ,A and B. (a) Identify any instant at which the two bicycles have the same velocity. (b) Which bicycle has the greater acceleration? (c) At what instant(s ) are the bicycles passing each other? (d) Which bicycle has the larger instantaneous velocity? (e) Which bicycle has the larger average velocity?

19. To find the magnitude of the car's acceleration, we can use the equation:<br /><br />$a = \frac{{v_f - v_i}}{{t}}$<br /><br />where $a$ is the acceleration, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time.<br /><br />In this case, the car comes to a stop, so the final velocity is 0. The initial velocity is given as 120 m / 5.0 s = 24 m/s. The time taken to come to a stop is 4.0 s.<br /><br />Plugging in the values, we get:<br /><br />$a = \frac{{0 - 24}}{{4}} = -6 \, \text{m/s}^2$<br /><br />The negative sign indicates that the acceleration is in the opposite direction of the initial velocity. Therefore, the magnitude of the acceleration is 6 m/s^2.<br /><br />To find the acceleration in g's, we divide the acceleration by the acceleration due to gravity, which is 9.80 m/s^2:<br /><br />$\text{Acceleration in g's} = \frac{{6}}{{9.80}} \approx 0.612 \, \text{g}$<br /><br />24. To find the length of the runway needed, we can use the equation:<br /><br />$d = v_i t + \frac{1}{2} a t^2$<br /><br />where $d$ is the distance, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.<br /><br />In this case, the initial velocity is 0 (since the plane starts from rest), the acceleration is 3.0 m/s^2, and the final velocity is 35 m/s.<br /><br />Plugging in the values, we get:<br /><br />$d = 0 \cdot t + \frac{1}{2} \cdot 3.0 \cdot t^2$<br /><br />Simplifying, we have:<br /><br />$d = 1.5 t^2$<br /><br />To find the time, we can rearrange the equation:<br /><br />$t = \sqrt{\frac{d}{1.5}}$<br /><br />Plugging in the value for the final velocity, we get:<br /><br />$t = \sqrt{\frac{35}{1.5}} \approx 5.92 \, \text{s}$<br /><br />Therefore, the length of the runway needed is:<br /><br />$d = 1.5 \cdot (5.92)^2 \approx 66.7 \, \text{m}$<br /><br />32. To determine if the driver will be able to stop in time, we can calculate the distance she needs to stop and compare it to the distance she has until the red light.<br /><br />First, let's calculate the distance she needs to stop. We can use the equation:<br /><br />$d = v_i t + \frac{1}{2} a t^2$<br /><br />where $d$ is the distance, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.<br /><br />In this case, the initial velocity is 18.0 m/s, the acceleration is -3.65 m/s^2 (since she is braking), and the time is 0.350 s.<br /><br />Plugging in the values, we get:<br /><br />$d = 18.0 \cdot 0.350 + \frac{1}{2} \cdot (-3.65) \cdot (0.350)^2$<br /><br />Simplifying, we have:<br /><br />$d = 6.3 - 0.438 \approx 5.762 \, \text{m}$<br /><br />Therefore, the driver needs to stop in 5.762 m.<br /><br />Now, let's calculate the distance she has until the red light. The distance is given as 20 m.<br /><br />Comparing the two distances, we see that the driver will not be able to stop in time, as she needs to stop in 5.762 m but has 20 m until the red light.<br /><br />To find the distance from the beginning of the intersection where she will be and in what direction, we can use the equation:<br /><br />$x = v_i t + \frac{1}{2} a t^2$<br /><br />where $x$ is the distance, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.<br /><br />Plugging in the values, we get:<br /><br />$x = 18.0 \cdot 0.350 + \frac{1}{2} \cdot (-3.65) \cdot (0.350)^2$<br /><br />Simplifying, we have:<br /><br />$x = 6.3 - 0.438 \approx 5.762 \, \text{m}$<br /><br />Therefore, she will be 5.762 m from the beginning of the intersection before the stop light.<br /><br />39. To find the height of the cliff, we can use the equation:<br /><br />$h = \frac{1}{2} g t^2$<br /><br />where $h$ is the