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A car is speeding up and has an instantaneous velocity of $1.0 \mathrm{m} / \mathrm{s}$ in the $+x$ -direction when a stopwatch reads $10.0 \mathrm{s} .$ It has a constant acceleration of $2.0 \mathrm{m} / \mathrm{s}^{2}$ in the $+x$ -direction. (a) What change in speed occurs between $t=10.0 \mathrm{s}$ and $t=12.0 \mathrm{s} ?(\mathrm{b})$ What is the speed when the stopwatch reads $12.0 \mathrm{s} ?$

(a) $\Delta v=4 \mathrm{m} / \mathrm{s}$(b) $v_{f}=5 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 2

Motion Along a Line

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

University of Michigan - Ann Arbor

Hope College

University of Winnipeg

McMaster University

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Okay, First, let's write what we know. We know that our initial velocity when it's measured at time, equals 10 seconds was one meter per second. And we also know that the car has a constant acceleration and that acceleration is equal to two meters per second squared. So for part A to calculate the change in speed, we're going to need the acceleration times, the change in time. So what is the change in time? Well, about the tea for this problem since we're going from 12 seconds to 10 seconds is the final minus. Initial times just two seconds. So using the numbers that we have here, we can say tell Ta VI or part A is equal to that two meters per second squared times our change in time, which is two seconds. It gives us a change in velocity of four meters per second. Okay, how for part B, What is the final speed? Well, the final speed for part B is equal to the initial speed, plus the change in the speed. So if we started at one meter per second and added four meters per second, our final answer is five meters per second

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