### Physics Problem: Comparing Distances under Constant AccelerationTom was running with his uncle. They both started from rest and ran with a constant positive acceleration the entire time. Both Tom and his uncle had the same acceleration. Tom’s uncle stopped at time 2t while Tom was still running. Tom ran until time 4t and then he stopped. Compared with his uncle, Tom ran a distance:- ○ one-half as far- ○ Twice as far- ○ 1.4 times as far- ○ four times as far- ○ five times as far### Explanation:The key concept to solve this problem involves understanding the relationship between time, acceleration, and distance traveled. For an object starting from rest and moving with constant acceleration, the distance \( s \) it travels is given by the formula:\[ s = \frac{1}{2} a t^2 \]Where \( a \) is the acceleration and \( t \) is the time.- **For Tom’s uncle**: He stopped at time \( 2t \), so the distance he traveled is: \[ s_{\text{uncle}} = \frac{1}{2} a (2t)^2 = 2a t^2 \]- **For Tom**: He ran until time \( 4t \), so the distance he traveled is: \[ s_{\text{Tom}} = \frac{1}{2} a (4t)^2 = 8a t^2 \]Therefore, compared with his uncle, Tom ran:\[ \text{Distance ratio} = \frac{s_{\text{Tom}}}{s_{\text{uncle}}} = \frac{8a t^2}{2a t^2} = 4 \]Tom ran four times as far as his uncle.

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Description: ### Physics Problem: Comparing Distances under Constant AccelerationTom was running with his uncle. They both started from rest and ran with a constant positive acceleration the entire time. Both Tom and his uncle had the same acceleration. Tom’s un

Given data: Initial velocity = u Constant acceleration = a Tom's uncle stopped in time (t1) = 2t…

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Tom was running with his uncle. They both started from rest and ran with a constant positive acceleration the entire time. Both Tom and his uncle had the same acceleration. Tom's uncle stopped at time 2t while Tom was still running. Tom ran until time 4t and then he stopped. Compared with his uncle, Tom ran a distance: one-half as far OTwice as far O 1.4 times as far O four times as far O five times as far

Tom was running with his uncle. They both started from rest and ran with a constant positive acceleration the entire time. Both Tom and his uncle had the same acceleration. Tom's uncle stopped at time 2t while Tom was still running. Tom ran until time 4t and then he stopped. Compared with his uncle, Tom ran a distance: one-half as far OTwice as far O 1.4 times as far O four times as far O five times as far

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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### Physics Problem: Comparing Distances under Constant Acceleration

Tom was running with his uncle. They both started from rest and ran with a constant positive acceleration the entire time. Both Tom and his uncle had the same acceleration. Tom’s uncle stopped at time 2t while Tom was still running. Tom ran until time 4t and then he stopped.

Compared with his uncle, Tom ran a distance:

- ○ one-half as far
- ○ Twice as far
- ○ 1.4 times as far
- ○ four times as far
- ○ five times as far

### Explanation:
The key concept to solve this problem involves understanding the relationship between time, acceleration, and distance traveled. For an object starting from rest and moving with constant acceleration, the distance \( s \) it travels is given by the formula:

\[ s = \frac{1}{2} a t^2 \]

Where \( a \) is the acceleration and \( t \) is the time.

- **For Tom’s uncle**: He stopped at time \( 2t \), so the distance he traveled is:
\[ s_{\text{uncle}} = \frac{1}{2} a (2t)^2 = 2a t^2 \]

- **For Tom**: He ran until time \( 4t \), so the distance he traveled is:
\[ s_{\text{Tom}} = \frac{1}{2} a (4t)^2 = 8a t^2 \]

Therefore, compared with his uncle, Tom ran:
\[ \text{Distance ratio} = \frac{s_{\text{Tom}}}{s_{\text{uncle}}} = \frac{8a t^2}{2a t^2} = 4 \]

Tom ran four times as far as his uncle.

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