Question

40 you are driving at 10.0 m/s when you suddenly slam on the brakes in 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.

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 formula for \( x \)

From \( v^2 = v_0^2 + 2ax \), 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 \( a = \frac{v - v_0}{t} \).

Step2: Substitute the values

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

1. Upward Motion: When the woman moves upward, her velocity decreases (since gravity acts downward). The velocity vectors (magnitude) get shorter. The change in velocity \( \Delta \vec{v} = \vec{v}_{\text{final}} - \vec{v}_{\text{initial}} \) points downward (because \( \vec{v}_{\text{final}} \) is smaller in magnitude and same direction, so \( \Delta \vec{v} = \text{smaller upward vector} - \text{larger upward vector} = \text{downward vector} \)). Acceleration \( \vec{a} = \frac{\Delta \vec{v}}{\Delta t} \), so \( \vec{a} \) is downward.
2. Downward Motion: When moving downward, velocity increases (gravity accelerates her downward). Velocity vectors (magnitude) get longer. \( \Delta \vec{v} = \vec{v}_{\text{final}} - \vec{v}_{\text{initial}} \) (both downward, final is larger). So \( \Delta \vec{v} \) is downward, hence \( \vec{a} = \frac{\Delta \vec{v}}{\Delta t} \) is downward.

Answer:

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

Problem 41