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Chapter 2: Problem 34

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.

Short Answer

Expert verified
(a) 285.7 meters. (b) 40 m/s. (c) & (d) Refer to the graph descriptions.

Step by step solution

01

Establish Car and Truck Equations

For the car, starting from rest with a constant acceleration, the motion equation is given by \( x_c = \frac{1}{2} a t^2 \), where \( a = 2.8 \text{ m/s}^2 \). Hence, \( x_c = 1.4 t^2 \).For the truck moving with a constant speed, its motion equation is \( x_t = v_t t \), where \( v_t = 20.0 \text{ m/s} \). Therefore, \( x_t = 20t \).
02

Set Equations Equal to Find Time

To find when the car overtakes the truck, set \( x_c = x_t \). This gives the equation \( 1.4t^2 = 20t \). Solve this by first dividing both sides by \( t \) (since \( t eq 0 \)) to obtain \( 1.4t = 20 \). Thus, \( t = \frac{20}{1.4} \approx 14.29 \text{ seconds} \).
03

Calculate Distance Overtaken

Substitute \( t = 14.29 \) back into any of the position equations, let's use \( x_c = 1.4 t^2 \):\[ x_c = 1.4 \times (14.29)^2 \approx 285.7 \text{ meters} \].Thus, the car overtakes the truck 285.7 meters beyond its starting point.
04

Calculate Speed of the Car

The car's velocity at the time it overtakes the truck is given by \( v_c = a t \).Using \( a = 2.80 \text{ m/s}^2 \) and \( t = 14.29 \text{ seconds} \):\[ v_c = 2.80 \times 14.29 \approx 40 \text{ m/s} \].
05

Draw the Distance-Time Graph

On the distance-time graph, plot the truck's path as a straight line and the car's path as a parabola originating from the origin. The intersection of these two curves indicates the position and time when the car overtakes the truck.
06

Draw the Velocity-Time Graph

On the velocity-time graph, represent the car's motion as a straight line with a positive slope (starting from zero) due to its constant acceleration. The truck's motion is represented as a horizontal line since its velocity is constant.

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Key Concepts

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

Constant Acceleration
In kinematics, one of the interesting scenarios is analyzing motion with constant acceleration. This means that the velocity of an object changes at a constant rate over time. For example, when a car accelerates from rest, it gradually increases its speed. If the car in our case accelerates at 2.80 m/s\(^2\), it gains speed consistently. The equation to describe the position of such an object over time is given by:
\[x = x_0 + v_0 t + \frac{1}{2} a t^2\]
For our scenario, the car starts from rest, simplifying it to \( x = \frac{1}{2} a t^2 \), as the initial velocity \( v_0 = 0 \). Constant acceleration equations are powerful tools for predicting motion in physics. They provide insights into how fast an object moves and the distance it covers over time.
Velocity-Time Graph
A velocity-time graph is a useful tool in understanding motion. It depicts how an object's speed changes over time. For our example, the velocity-time graph of the car, which has constant acceleration, starts at the origin and ascends in a straight line. This straight line signifies a steady increase in velocity.
  • The slope of this line represents acceleration鈥攕teeper slopes indicate higher acceleration.
  • For the car with 2.80 m/s\(^2\) acceleration, its graph is a straight line with a positive slope.

Conversely, the truck moves at a constant speed of 20.0 m/s. Its velocity-time graph is a horizontal line, meaning its speed doesn鈥檛 change over time. The differences in these representations make it easier to understand how different vehicles behave over the same period.
Motion Equations
Motion equations are essential in solving kinematics problems. They link displacement, velocity, acceleration, and time. One of the fundamental equations when dealing with constant acceleration is:
\[v = v_0 + at\]
It helps in finding the velocity of an object at any given time. In our context, this equation tells us the car鈥檚 velocity when it eventually overtakes the truck.
  • By plugging in the values, we find that the car's velocity when it catches up to the truck is approximately 40 m/s.
  • This is derived using its constant acceleration and the time it has been accelerating.

These equations form the bedrock of analyzing motion scenarios, allowing us to calculate various unknowns given some known variables.
Distance-Time Graph
Distance-time graphs are visual representations of an object's position over time. They reveal how far an object travels as time progresses. For the car accelerating away from rest, the graph is a curve opening upward, a parabolic shape due to the quadratic nature of the equation \( x = \frac{1}{2}at^2 \).
  • The curve starts at the origin and its steepness increases, representing the car's increasing speed over time.
  • At any given point, the slope of the curve corresponds to the car's velocity at that time.

The truck, however, travels at a constant speed, so its distance-time graph is a straight, diagonal line starting from the intersection. The point where the car鈥檚 curve intersects the truck鈥檚 line is the exact location and time where the car overtakes the truck. This intersection is key in understanding relative motion between the two vehicles.

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Most popular questions from this chapter

A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler's hand?

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. Ignore air resistance. (a) If the height of the building is 20.0 m, what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_0\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_0\) is 6.0 m/s and (ii) \(v_0\) is 9.5 m/s? (c) If \(v_0\) is greater than some value \(v_{max}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{max}\). The value \(v_{max}\) has a simple physical interpretation. What is it? (d) If \(v_0\) is less than some value \(v_{min}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{min}\). The value \(v_{min}\) also has a simple physical interpretation. What is it?

If the contraction of the left ventricle lasts 250 ms and the speed of blood flow in the aorta (the large artery leaving the heart) is 0.80 m/s at the end of the contraction, what is the average acceleration of a red blood cell as it leaves the heart? (a) 310 ms\(^2\); (b) 31 m/s\(^2\); (c) 3.2 m/s\(^2\); (d) 0.32 m/s\(^2\).

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.

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?
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