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Ch 02: Kinematics in One Dimension
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 2, Problem 68b

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s² at the instant when David passes. What is her speed as she passes him?

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Step 1: Define the motion equations for both David and Tina. David is moving at a constant velocity, so his position as a function of time is given by: xd=30t. Tina starts from rest and accelerates uniformly, so her position as a function of time is: xt=12at2, where a=2.0 m/s².
Step 2: Set the positions of David and Tina equal to each other to find the time t when Tina passes David. This gives the equation: 30t=122.0t2.
Step 3: Simplify the equation to solve for t. Rearrange to get: t2=301t. Solve for t by factoring or using algebraic methods.
Step 4: Once you have the time t, calculate Tina's velocity at that moment using the formula for velocity under constant acceleration: v=u+at, where u=0 (initial velocity) and a=2.0 m/s².
Step 5: Substitute the value of t into the velocity equation to find Tina's speed as she passes David. This will give you the final answer.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Relative Motion

Relative motion refers to the calculation of the motion of an object as observed from a particular reference point. In this scenario, David is moving at a constant speed while Tina is initially at rest. Understanding relative motion is crucial to determine how fast Tina needs to accelerate to catch up to David.
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Intro to Relative Motion (Relative Velocity)

Acceleration

Acceleration is the rate of change of velocity of an object with respect to time. In this case, Tina accelerates at a steady rate of 2.0 m/s². This concept is essential for calculating how her speed increases over time as she starts from rest.
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Intro to Acceleration

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. They relate displacement, initial velocity, final velocity, acceleration, and time. To find Tina's speed as she passes David, we can use these equations to determine her final velocity after a certain time of acceleration.
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Related Practice
Textbook Question

When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first 20 s is extremely well modeled by the simple equation vx2 = (2P/m)t, where P = 3.6 ✕ 10⁴ watts is the car's power output, m = 1200 kg is its mass, and vx is in m/s. That is, the square of the car's velocity increases linearly with time. Find an algebraic expression in terms of P, m, and t for the car's acceleration at time t.

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Textbook Question

Nicole throws a ball straight up. Chad watches the ball from a window 5.0 m above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of 10 m/s as it passes him on the way back down. How fast did Nicole throw the ball?

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Textbook Question

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500m/s . It is loaded with 600 kg of fuel, which it burns in 30 s. What is the rocket's speed 10 s, 20 s, and 30 s after launch?

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Textbook Question

A motorist is driving at 20 m/s when she sees that a traffic light 200 m ahead has just turned red. She knows that this light stays red for 15 s, and she wants to reach the light just as it turns green again. It takes her 1.0 s to step on the brakes and begin slowing. What is her speed as she reaches the light at the instant it turns green?

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Textbook Question

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s² at the instant when David passes. How far does Tina drive before passing David?

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Textbook Question

If a Tesla Model S P100D in 'Ludicrous mode' is pushed to its limit, the first 3.0 s3.0\(\text{ s}\) of acceleration can be modeled as

ax={(35m/s3)t0 st0.40s14.6m/s2(1.5m/s3)t0.40st3.0sa_{x}=\(\begin{cases}\)(35\,\(\text{m/s}\)^3)t & 0\(\text{ s}\]\le\) t\(\le\)0.40\,\(\text{s}\)\\ 14.6\,\(\text{m/s}\)^2-(1.5\,\(\text{m/s}\)^3)t & 0.40\,\(\text{s}\[\le\) t\(\le\)3.0\,\(\text{s}\]\end{cases}\)

What acceleration would be needed to achieve the same speed in the same time at constant acceleration? Give your answer as a multiple of gg.

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