Question

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

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

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 \(d_{1}\).

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.

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.

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:

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.

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).

Key concepts

Constant Acceleration

Constant acceleration means that the rate at which an object's velocity changes is steady over time.
This is crucial to solve problems in kinematics effectively, as acceleration typically determines how quickly an object speeds up or slows down.
When dealing with constant acceleration, there are specific formulas that can help calculate distances and velocities.

Here are a few key points about constant acceleration:

• The formula for distance (\(d\)) covered under constant acceleration is \(d = ut + \frac{1}{2}at^2\), where \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
• If the initial velocity \(u\) is zero, as in the exercise where the girl starts accelerating from rest, the formula simplifies to \(d = \frac{1}{2}at^2\).

The girl's acceleration of \(2.2 \mathrm{~m/s^2}\) is what enables her to "catch up" to her friend over time. By understanding and using the constant acceleration formula, we can determine how long it will take her to meet her friend.

Constant Speed

An object moving at constant speed covers equal distances in equal intervals of time.
This implies that there is no acceleration involved, meaning the velocity doesn’t change at all.

Key aspects of constant speed include:

• Speed (\(v\)) remains the same throughout the motion.
• The formula for distance in this case is \(d = vt\), where \(d\) is the distance, \(v\) is the constant speed, and \(t\) is the time travelled.

In the exercise, the friend passes by the girl while traveling at a constant speed of \(8.0 \mathrm{~m/s}\).
This constant motion allows us to calculate the distance travelled during the 20 seconds when the girl stops. Understanding this concept is critical because it sets up the initial distance gap the girl needs to close to catch her friend.

Distance Calculation

Calculating distance in kinematics can vary depending on whether the movement involves constant speed or constant acceleration.
The exercise involves both these concepts and requires careful application of appropriate formulas.

For constant speed, the formula \(d = vt\) calculates the distance covered in a given time.
This is straightforward when the speed doesn’t change.
The friend's distance after 20 seconds, when moving at \(8.0 \mathrm{~m/s}\), can directly be calculated using this formula.

When considering constant acceleration, we use the formula \(d = ut + \frac{1}{2}at^2\).
This accounts for the change in speed over time and provides the distance covered during acceleration.
The girl's distance from the point she starts pedaling can be calculated this way.
To find out when she catches her friend, this distance becomes crucial.By setting the calculated distances equal, we solve for the time \(t\) when the girl meets her friend. Knowing how to use these formulas correctly allows us to predict and understand motion scenarios.