/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 A car and a truck start from res... [FREE SOLUTION] | 91Ó°ÊÓ

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A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of 2.10 m/s\(^2\), and the car has an acceleration of 3.40 m/s\(^2\). The car overtakes the truck after the truck has moved 60.0 m. (a) How much time does it take the car to overtake the truck? (b) How far was the car behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x =\) 0 at the initial location of the truck.

Short Answer

Expert verified
It takes 10.67 seconds for the car to overtake the truck, initially 19.2 meters behind. Their speeds are 22.4 m/s (truck) and 36.3 m/s (car) when abreast.

Step by step solution

01

Understanding the Problem

Let's break down what is given: the truck starts with an acceleration of 2.10 m/s² and moves 60.0 m, while the car starts from an unknown initial position with an acceleration of 3.40 m/s². We need to find the time it takes for the car to catch up with the truck, the initial distance between them, and their speeds when they are at the same position, and also sketch their positions over time.
02

Equation of Motion for the Truck

Use the equation of motion for the truck: \( x_t = \frac{1}{2} a_t t^2 \). Here, \( a_t = 2.10 \, \text{m/s}^2 \) and \( x_t = 60.0 \, \text{m} \). Thus, \( 60.0 = \frac{1}{2} \times 2.10 \times t^2 \). Simplify to find \( t \).
03

Solving for Time (Truck)

Solving \( 60.0 = 1.05 t^2 \) gives us \( t = \sqrt{\frac{60}{1.05}} \). Calculate this to find \( t \approx 10.67 \text{ seconds} \).
04

Equation of Motion for the Car

Use the same form of equation of motion for the car but consider an initial distance \( d \): \( x_c = d + \frac{1}{2} a_c t^2 \), where \( a_c = 3.40 \, \text{m/s}^2 \). When the car overtakes the truck, \( x_c = x_t = 60 \), and the time \( t \) is the same.
05

Solving for the Initial Distance

Set \( 60 = d + \frac{1}{2} \times 3.40 \times (10.67)^2 \). Solve for \( d \) to find \( d \approx 19.2 \, \text{m} \), meaning the car was initially 19.2 meters behind the truck.
06

Finding the Velocity of the Truck

Use the formula \( v_t = a_t t \) for the truck. Substitute \( a_t = 2.10 \) and \( t = 10.67 \) to find the velocity: \( v_t = 2.10 \times 10.67 = 22.4 \, \text{m/s} \).
07

Finding the Velocity of the Car

Use a similar formula \( v_c = a_c t \) for the car. Substitute \( a_c = 3.40 \) and \( t = 10.67 \) to find \( v_c = 3.40 \times 10.67 = 36.3 \, \text{m/s} \).
08

Sketching the Graphs

Plot the position vs. time for both the car and the truck. Start the truck at \((0, 0)\) and follow its quadratic trajectory \( x = 1.05 t^2 \). Start the car at \( (0, -19.2) \) and follow \( x = -19.2 + 1.7 t^2 \). Both should intersect at \( (10.67, 60) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Acceleration
Understanding uniform acceleration is key to solving kinematics problems like the one in our exercise. When an object accelerates uniformly, it means that the rate of change of its velocity is constant throughout its motion. This concept simplifies calculations since the acceleration value remains unchanged over time.

In the case of our truck and car problem, both the car and the truck start from rest, implying their initial velocity is zero. They accelerate at different constant rates of 2.10 m/s two for the truck and 3.40 m/s two for the car. This constant acceleration means we can reliably apply kinematic equations to predict their future positions and velocities based on time elapsed.

Uniform acceleration also allows us to use straightforward formulas, like the equations of motion and velocity-time graphs, to understand and solve the problem effectively.
Equations of Motion
The equations of motion are essential tools when dealing with problems involving uniform acceleration. These equations help relate the motion characteristics of an object, like displacement, time, initial velocity, and acceleration.

In our exercise, we apply one of the most fundamental equations of motion: \[ x = ut + \frac{1}{2} a t^2 \] Where:
  • \( x \) is the displacement
  • \( u \) is the initial velocity, which is zero for both the car and the truck
  • \( a \) is the constant acceleration
  • \( t \) is the time elapsed

For the truck, substituting into this equation allows us to find the time it takes to travel 60 meters. Similarly, for the car, this equation helps determine its position relative to the truck's position when both vehicles are side by side.

We also used the equation \( v = u + at \) to calculate the velocities of the car and truck afterward, which tells us how fast each was moving at the point of overtaking.
Velocity-time Graph
A velocity-time graph is a powerful visual tool that allows us to understand how the velocity of an object changes over time. It provides insights into the nature of an object's motion, especially when dealing with uniformly accelerated motion.

For objects with constant acceleration, like our car and truck, the velocity-time graph is a straight line. The slope of this line represents the acceleration. In this scenario, a steeper slope indicates a higher acceleration, which is why the car's line is steeper than the truck's.

The area under the velocity-time graph gives the displacement. For our problem, this information helps verify the calculations made using equations of motion. In essence, plotting these graphs for both vehicles, we can see how each progresses over time and pinpoint the moment they are side by side, reinforcing the algebraic solution with a visual check.

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Most popular questions from this chapter

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 2.80 m/s\(^2\). At the same instant a truck, traveling with a constant speed of 20.0 m/s, overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car traveling when it overtakes the truck? (c) Sketch an \(x-t\) graph of the motion of both vehicles. Take \(x=\) 0 at the intersection. (d) Sketch a \(v_x-t\) graph of the motion of both vehicles.

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 m/s\(^2\) upward. At 25.0 s after launch, the second stage fires for 10.0 s, which boosts the rocket's velocity to 132.5 m/s upward at 35.0 s after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stage-two rocket be moving just as it reaches the launch pad?

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 200 m during the second 5.0 s of its motion. How far did it roll during the first 5.0 s of motion?

The fastest measured pitched baseball left the pitcher's hand at a speed of 45.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. When you see the meter stick released, you grab it with those two fingers. You can calculate your reaction time from the distance the meter stick falls, read directly from the point where your fingers grabbed it. (a) Derive a relationship for your reaction time in terms of this measured distance, \(d\). (b) If the measured distance is 17.6 cm, what is your reaction time?

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