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Chapter 2: Problem 32

An object starts from rest at \(x=0\) m at time \(t=0\) s. Five seconds later, at \(t=5.0 \mathrm{s},\) the object is observed to be at \(x=40.0 \mathrm{m}\) and to have velocity \(v_{x}=11 \mathrm{m} / \mathrm{s}\) a. Was the object's acceleration uniform or nonuniform? Explain your reasoning. b. Sketch the velocity-versus-time graph implied by these data. Is the graph a straight line or curved? If curved, is it concave upward or downward?

Short Answer

Expert verified
a. The object's acceleration is uniform, as the velocity changes linearly over time. b. The velocity versus time graph is a straight line, neither concave up nor downward.

Step by step solution

01

Interpret the given data

We know that at time \(t=0\) s, the object is at rest, i.e., its initial velocity \(u = 0\) m/s. At \(t=5.0\) s, its position \(x=40.0\) m and velocity \(v_{x}=11\) m/s.
02

Determine the type of acceleration

To find out if an object's acceleration is uniform or non-uniform, we calculate the average acceleration between \(t=0\) s and \(t=5.0\) s using the formula \(\overline{a}=\frac{v-u}{t}\). Here \(v=11\) m/s, \(u=0\) m/s, and \(t=5.0\) s. After insertion, \(\overline{a}=2.2\) m/s虏. As the velocity changes uniformly over time, it suggests that acceleration is constant.
03

Sketch velocity versus time graph

We plot the velocity versus time graph by marking the given points (0,0) and (5,11). The velocity changes linearly over time i.e., the velocity-time graph is a straight line. The line is neither concave up nor downtward.

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Key Concepts

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

Uniform Acceleration
In the world of kinematics, understanding acceleration is key. Acceleration tells us how quickly an object speeds up or slows down. When we talk about uniform acceleration, it means the object's velocity changes at a constant rate over time. In simpler terms, the acceleration does not vary.

In this exercise, we see that the object starts from rest and achieves a certain velocity over a specified time period. The calculation shows that average acceleration is 2.2 m/s虏. Since this average acceleration is consistent with the data provided (where the velocity increases uniformly from 0 to 11 m/s in 5 seconds), the object's acceleration can be classified as uniform. This implies no sudden changes in motion, leading to smoother and predictable movement.
Velocity-Time Graph
A velocity-time graph is a powerful tool for visually understanding motion. It shows how velocity changes over time, helping us grasp the object's acceleration pattern.

In this scenario, we're asked to sketch a velocity-time graph based on the given data points. We have the initial point at (0,0) and another at (5,11), representing time in seconds and velocity in meters per second, respectively. Connecting these points results in a straight line.

A straight line indicates uniform acceleration. The slope of the line reflects the rate of acceleration, which is constant in this case. There is no curve or concavity, further confirming uniform acceleration.
Average Acceleration
Average acceleration provides a simple way to measure how velocity changes over a given period. It is computed using the formula \( \overline{a} = \frac{v - u}{t} \), where \(v\) is final velocity, \(u\) is initial velocity, and \(t\) is time.

Here, the initial velocity \(u\) is 0 m/s, the final velocity \(v\) is 11 m/s, and the time \(t\) is 5 seconds. By plugging these into the formula, we find \( \overline{a} = 2.2 \) m/s虏. This average acceleration is critical in confirming whether the object's acceleration was uniform.

A consistent value for average acceleration, as calculated, supports the conclusion that the motion was steady and predictable. Understanding average acceleration helps highlight whether motion conditions remain the same over time or vary significantly.

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Most popular questions from this chapter

A ball is thrown vertically upward with a speed of \(19.6 \mathrm{m} / \mathrm{s}\) a. What is the ball's velocity and its height after 1.0,2.0,3.0 and \(4.0 \mathrm{s} ?\) b. Draw the ball's velocity-versus-time graph. Give both axes an appropriate numerical scale.

A car starts from rest at a stop sign. It accelerates at \(4.0 \mathrm{m} / \mathrm{s}^{2}\) for \(6.0 \mathrm{s},\) coasts for \(2.0 \mathrm{s},\) and then slows down at a rate of \(3.0 \mathrm{m} / \mathrm{s}^{2}\) for the next stop sign. How far apart are the stop signs?

The minimum stopping distance for a car traveling at a speed of \(30 \mathrm{m} / \mathrm{s}\) is \(60 \mathrm{m},\) including the distance traveled during the driver's reaction time of \(0.50 \mathrm{s}\) a. What is the minimum stopping distance for the same car traveling at a speed of \(40 \mathrm{m} / \mathrm{s} ?\) b. Draw a position-versus-time graph for the motion of the car in part a. Assume the car is at \(x_{0}=0\) m when the driver first sees the emergency situation ahead that calls for a rapid halt.

A rubber ball is shot straight up from the ground with speed \(v_{0}\) Simultaneously, a second rubber ball at height \(h\) directly above the first ball is dropped from rest. a. At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of \(v_{0}\) and \(g\). b. What is the maximum value of \(h\) for which a collision occurs before the first ball falls back to the ground? c. For what value of \(h\) does the collision occur at the instant when the first ball is at its highest point?

A driver has a reaction time of \(0.50 \mathrm{s}\), and the maximum deceleration of her car is \(6.0 \mathrm{m} / \mathrm{s}^{2} .\) She is driving at \(20 \mathrm{m} / \mathrm{s}\) when suddenly she sees an obstacle in the road \(50 \mathrm{m}\) in front of her. Can she stop the car in time to avoid a collision?
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