Question

39 at t = 0 s, a ball is launched straight upward with a velocity of 19.6 m/s. the ball slows with constant acceleration until it stops at t = 2.00 s. the ball falls back to the ground with the same acceleration until it hits the ground with velocity -19.6 m/s at t = 4.00 s. what is the ball’s total displacement after 2.00 s? what is the ball’s acceleration during its entire motion?
40 you are driving at 10.0 m/s when you suddenly slam on the brakes to avoid a collision. you come to a complete stop. when you stop, your acceleration is determined by your tires and the road conditions, and it is a constant -4.1 m/s². the skid marks you leave behind can be measured to determine your displacement. how long will your skid marks be?
41 to determine a car’s maximum acceleration, you start from rest, press down hard on the accelerator pedal, and time how long it takes to get up to 26.8 m/s (about 60 miles per hour). it takes a powerful new electric car only 3.50 s to reach 26.8 m/s. what is the car’s acceleration?
42 sep develop a model using the velocity vectors from the dot diagram for a woman jumping on a trampoline, show that the acceleration vector is downward for both her upward and downward motions.
43 your physics teacher crumples a piece of paper into a ball and releases it from rest 1.5 m from the ground. how long does the ball take to hit the ground?

Explanation:

Problem 40:

Step1: Identify known variables

Initial velocity \( v_0 = 10.0 \, \text{m/s} \), final velocity \( v = 0 \, \text{m/s} \) (comes to stop), acceleration \( a = -4.1 \, \text{m/s}^2 \). We use the kinematic equation \( v^2 = v_0^2 + 2ax \) to find displacement \( x \).

Step2: Rearrange the equation for \( x \)

From \( v^2 = v_0^2 + 2ax \), we solve for \( x \):
\( x = \frac{v^2 - v_0^2}{2a} \)

Step3: Substitute the values

Substitute \( v = 0 \), \( v_0 = 10.0 \, \text{m/s} \), \( a = -4.1 \, \text{m/s}^2 \):
\( x = \frac{0 - (10.0)^2}{2(-4.1)} = \frac{-100}{-8.2} \approx 12.2 \, \text{m} \)

Step1: Identify known variables

Initial velocity \( v_0 = 0 \, \text{m/s} \) (starts from rest), final velocity \( v = 26.8 \, \text{m/s} \), time \( t = 3.50 \, \text{s} \). Use the kinematic equation \( v = v_0 + at \) to find acceleration \( a \).

Step2: Rearrange the equation for \( a \)

From \( v = v_0 + at \), solve for \( a \):
\( a = \frac{v - v_0}{t} \)

Step3: Substitute the values

Substitute \( v = 26.8 \, \text{m/s} \), \( v_0 = 0 \), \( t = 3.50 \, \text{s} \):
\( a = \frac{26.8 - 0}{3.50} \approx 7.66 \, \text{m/s}^2 \)

Step1: Identify known variables

Initial velocity \( v_0 = 0 \, \text{m/s} \) (released from rest), displacement \( y = 1.5 \, \text{m} \), acceleration \( a = g = 9.8 \, \text{m/s}^2 \). Use the kinematic equation \( y = v_0t+\frac{1}{2}at^2 \).

Step2: Simplify the equation (since \( v_0 = 0 \))

The equation becomes \( y=\frac{1}{2}gt^2 \). Solve for \( t \):
\( t = \sqrt{\frac{2y}{g}} \)

Step3: Substitute the values

Substitute \( y = 1.5 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \):
\( t=\sqrt{\frac{2\times1.5}{9.8}}=\sqrt{\frac{3}{9.8}}\approx\sqrt{0.306}\approx0.553 \, \text{s} \)

Answer:

The length of the skid marks is approximately \( \boldsymbol{12.2 \, \text{meters}} \).

Problem 41: