Practice Test Questions: Truck Acceleration & 2-Block Friction System Analysis

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In summary, the first conversation involves a truck passing a red light at a speed of 20 m/s and accelerating at 0.15 m/s/s, with a police car starting from rest 30 m behind with acceleration 0.6 m/s/s to pursue the truck. The distance from the traffic light when the police caught up with the truck is to be found, and the position versus time information for both vehicles will be plotted on the same graph. If the catch-up distance is doubled, the truck acceleration will also be found.The second conversation involves a 2-block friction system with masses of 2 kg and 10 kg, a coefficient of friction of 0.25, and an angle of 30 degrees. The
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xxpundoxx
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1.A truck passed a red light at a speed of 20 m/s and started accelerating at 0.15 m/s/s. A police car 30 m behind the truck started from rest with acceleration 0.6 m/s/s to pursuit the truck. Find the distance from the traffic light when the police caught up with the truck. Plot the position versus time information for the police car and truck on the same graph. Find the truck acceleration if the catch-up distance is double.

2.Given the following 2-block friction system, does the system move? Show calculation
If it moves, calculate the tension and the acceleration.
a=2 kg, b=10 kg, u=.25, angle=30 degrees. find the acceleration and tension
T-(2)(9.8) sin30 - (0.25)(2)(9.8) cos30 = 2a
 
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Welcome to PF xxpundoxx.
I assume you just missed the big bold letters that ask you to use the template and provide some work you have already tried, or at least write down some formulas you think might be relevant?

Anyway, the question asks you to plot the position versus time graph of both the police car and the truck in the same graph. Before doing any calculation: how would this help you find the answer? Now give me a formula you are going to use to solve the question.
 
  • #3


I would approach these questions by first analyzing the given information and identifying the relevant equations and principles that can be applied.

For the first question, we can use the equations of motion to calculate the distance traveled by the truck and police car. The initial position of the truck is not given, so we can assume it is at the traffic light (x=0) when it passes the red light. The equation for distance traveled (x) as a function of time (t) is x = x0 + v0t + 1/2at^2, where x0 is the initial position, v0 is the initial velocity, and a is the acceleration. For the truck, x0=0, v0=20 m/s, and a=0.15 m/s^2. Using this equation, we can calculate the distance traveled by the truck when the police car starts pursuing it:

x = (20 m/s)(t) + 1/2(0.15 m/s^2)(t^2)

Next, we can use the same equation for the police car, with x0=30 m (since it starts 30 m behind the truck), v0=0, and a=0.6 m/s^2. We want to find the time (t) when the police car catches up with the truck, which is when their positions are equal. Setting the two equations for x equal to each other, we have:

(20 m/s)(t) + 1/2(0.15 m/s^2)(t^2) = (30 m) + (1/2)(0.6 m/s^2)(t^2)

Solving for t, we get t=40 s. So the police car catches up with the truck after 40 seconds. To find the distance from the traffic light when this happens, we plug t=40 s into either equation for x:

x = (20 m/s)(40 s) + 1/2(0.15 m/s^2)(40 s)^2 = 800 m + 240 m = 1040 m

To plot the position versus time information for both the truck and police car on the same graph, we can use the equations we found for x as a function of t. The graph would show the truck starting at x=0 and moving with a constant positive slope (since its velocity is positive
 

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