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Chapter 2: Problem 65

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?

Short Answer

Expert verified
b) What is the constant acceleration the girl would need if she had been rolling along at \(1.2 \mathrm{~m} / \mathrm{s}\) when her friend passed, so she can catch up with her friend in the same amount of time?

Step by step solution

01

Calculate the distance traveled by her friend during the 20 seconds the girl is taking a break.

Using the formula for distance under constant speed: $$ d_{1} = v_{1}t $$ where \(v_{1}=8.0 \mathrm{~m} / \mathrm{s}\) is the friend's speed and \(t=20 \mathrm{s}\) is the time the girl takes a break. Calculate \(d_{1}\).
02

Calculate the distance traveled by the girl with constant acceleration.

Using the formula for distance under constant acceleration: $$ d_{2} = ut + \frac{1}{2}at^2 $$ where \(u=0 \mathrm{~m} / \mathrm{s}\) is the girl's initial speed, \(a=2.2 \mathrm{~m} / \mathrm{s}^{2}\) is her acceleration, and \(t\) is the time for her to catch up with her friend.
03

Calculate the total distance that the friend travels before the girl catches up to her.

We know that the friend's speed is constant and that the girl needs to travel the same distance to catch up to her friend. Therefore, using the distance formula for constant speed: $$ d_{3} = v_{1}(20 + t) $$ where \(20+t\) is the total time for the friend before the girl catches up to her.
04

Set the distances equal to each other and solve for t.

In order to catch up with her friend, the girl needs to travel the same distance as her friend (\(d_{2} = d_{3}\)). Therefore: $$ ut + \frac{1}{2}at^2 = v_{1}(20 + t) $$ Plug in the values and solve for \(t\). b) To find the constant acceleration needed by the girl if she had been rolling along at \(1.2 \mathrm{~m} / \mathrm{s}\) when her friend passed, so she can catch up with her friend in the same amount of time:
05

Calculate the distance needed to be covered in order to catch up with her friend.

Using the results from part (a), we know the distance that needs to be covered by the girl in order to catch up with her friend.
06

Calculate the acceleration needed using the same formula for distance under constant acceleration.

This time, we know the time, initial velocity, and the distance that needs to be covered by the girl: $$ d = ut + \frac{1}{2}at^2 $$ Plug in the values and solve for \(a\), the acceleration needed for the girl to catch her friend in the same amount of time as in part (a).

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