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Chapter 2: Problem 55

A car starts from rest and accelerates at \(10.0 \mathrm{~m} / \mathrm{s}^{2}\) How far does it travel in 2.00 s?

Short Answer

Expert verified
Answer: The car travels a distance of 20.0 meters in 2.00 seconds.

Step by step solution

01

List given information

The problem provides us with the following information: - The car starts from rest, meaning its initial velocity is 0: \(v_0 = 0 \mathrm{~m}/\mathrm{s}\) - The car accelerates at a constant rate: \(a = 10.0 \mathrm{~m}/\mathrm{s}^{2}\) - We need to find the distance traveled in a given time: \(t = 2.00 \mathrm{~s}\)
02

Use the kinematic equation to determine the distance traveled

Since we have the initial velocity, acceleration, and time, we can use the following kinematic equation to find the distance traveled: $$ d = v_0t + \frac{1}{2}at^2 $$ This equation states that the distance traveled (d) is equal to the product of the initial velocity (v_0) and time (t) plus one-half the product of the acceleration (a) and the square of the time (t^2).
03

Substitute the given values and solve for the distance

Substitute the given values into the equation: $$ d = (0 \mathrm{~m}/\mathrm{s})(2.00 \mathrm{~s}) + \frac{1}{2}(10.0 \mathrm{~m}/\mathrm{s}^{2})(2.00 \mathrm{~s})^2 $$ Now, simplify the equation and solve for the distance: $$ d = 0 + (10.0 \mathrm{~m}/\mathrm{s}^{2})(4.00 \mathrm{~s}^2)/2 $$ $$ d = (20.0 \mathrm{~m}/\mathrm{s}^{2})(2.00 \mathrm{~s}^2) $$ $$ d = 20.0 \mathrm{~m} $$
04

State the final answer

The car travels a distance of \(20.0\) meters in \(2.00\) seconds.

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Most popular questions from this chapter

A ball is dropped from the roof of a building. It hits the ground and it is caught at its original height 5.0 s later. a) What was the speed of the ball just before it hits the ground? b) How tall was the building? c) You are watching from a window \(2.5 \mathrm{~m}\) above the ground. The window opening is \(1.2 \mathrm{~m}\) from the top to the bottom. At what time after the ball was dropped did you first see the ball in the window?

A ball is thrown directly downward, with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\), from a height of \(50.0 \mathrm{~m}\). After what time interval does the ball strike the ground?

The position of a particle moving along the \(x\) -axis is given by \(x=\left(11+14 t-2.0 t^{2}\right),\) where \(t\) is in seconds and \(x\) is in meters. What is the average velocity during the time interval from \(t=1.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s} ?\)

A girl is riding her bicycle. When she gets to a corner, she stops to get a drink from her water bottle. At that time, a friend passes by her, traveling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). a) After \(20 \mathrm{~s}\), the girl gets back on her bike and travels with a constant acceleration of \(2.2 \mathrm{~m} / \mathrm{s}^{2} .\) How long does it take for her to catch up with her friend? b) If the girl had been on her bike and rolling along at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\) when her friend passed, what constant acceleration would she need to catch up with her friend in the same amount of time?

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How high above the projection point is it after 1.00 s?
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