Bob is cruising leisurely on his bike at 2.5 m/s when Jim passes him, traveling at a constant 9.8 m/s. Bob immediately begins accelerating at 2.8 m/s² until he reaches his maximum velocity of 13.3 m/s. He continues cruising at that speed until he catches Jim.

How long did it take him to catch up to Jim? (Assume the clock started at the instant he began to accelerate

To determine how long it took Bob to catch up to Jim, we can break this problem into two phases:

1. **Phase 1:** Bob accelerates from 2.5 m/s to his maximum velocity of 13.3 m/s.
2. **Phase 2:** Bob maintains a constant velocity of 13.3 m/s and catches up to Jim, who is moving at 9.8 m/s.

### Phase 1: Acceleration to 13.3 m/s

- Initial velocity \(v_0 = 2.5 \, \text{m/s}\)
- Final velocity \(v_f = 13.3 \, \text{m/s}\)
- Acceleration \(a = 2.8 \, \text{m/s}^2\)

Using the equation for motion under constant acceleration:

\[
v_f = v_0 + a t_1
\]

Solving for \(t_1\) (the time to reach 13.3 m/s):

\[
t_1 = \frac{v_f - v_0}{a}
\]

Substituting the values:

\[
t_1 = \frac{13.3 - 2.5}{2.8}
\]

Let's calculate this value.

\[
t_1 = \frac{10.8}{2.8} = 3.857 \, \text{seconds}
\]

So, it takes Bob approximately 3.857 seconds to reach his maximum speed of 13.3 m/s.

Now, let's calculate the distance Bob travels during this time using the equation:

\[
d_1 = v_0 t_1 + \frac{1}{2} a t_1^2
\]

Substituting the values:

\[
d_1 = 2.5 \times 3.857 + \frac{1}{2} \times 2.8 \times (3.857)^2
\]

Now, we calculate \(d_1\).

\[
d_1 = 9.6425 + 0.5 \times 2.8 \times 14.873
\]

\[
d_1 = 9.6425 + 20.222
\]

\[
d_1 = 29.8645 \, \text{meters}
\]

### Phase 2: Constant speed

After reaching 13.3 m/s, Bob continues at this speed until he catches up to Jim.

In the same time \(t_1 = 3.857 \, \text{seconds}\), Jim has traveled:

\[
d_J = v_J t_1 = 9.8 \times 3.857 = 37.793 \, \text{meters}
\]

Thus, by the time Bob reaches his maximum velocity, he is behind by:

\[
\text{Distance behind} = d_J - d_1 = 37.793 - 29.8645 = 7.9285 \, \text{meters}
\]

Now Bob and Jim are both moving at constant velocities, with Bob traveling at 13.3 m/s and Jim at 9.8 m/s. The relative speed between Bob and Jim is:

\[
\Delta v = 13.3 - 9.8 = 3.5 \, \text{m/s}
\]

The time it takes for Bob to cover the remaining 7.9285 meters is:

\[
t_2 = \frac{\text{Distance behind}}{\Delta v} = \frac{7.9285}{3.5} = 2.265 \, \text{seconds}
\]

### Total Time

The total time for Bob to catch Jim is the sum of \(t_1\) and \(t_2\):

\[
t_{\text{total}} = t_1 + t_2 = 3.857 + 2.265 = 6.122 \, \text{seconds}
\]

Thus, it took Bob approximately **6.12 seconds** to catch up to Jim.

Would you like more details on any part of the solution?

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Here are 5 related questions to expand on this:

1. How far did Jim travel by the time Bob caught up to him?
2. What would the time be if Bob's maximum velocity were lower?
3. How much distance did Bob travel in total to catch Jim?
4. What is the relative acceleration of Bob compared to Jim?
5. What happens if Jim accelerates instead of moving at a constant speed?

**Tip:** To solve problems involving multiple phases, break them down into smaller parts and use relevant equations for each phase.

Bob is cruising leisurely on his bike at 2.5 m/s when Jim passes him, traveling at a constant 9.8 m/s. Bob immediately begins accelerating at 2.8 m/s² until he reaches his maximum velocity of 13.3 m/s. He continues cruising at that speed until he catches Jim. How long did it take him to catch up to Jim? (Assume the clock started at the instant he began to accelerate).

Solve a kinematics problem where Bob accelerates to catch up with Jim, who is traveling at a constant speed. Learn to apply equations of motion under constant acceleration and relative velocity to determine the time it takes Bob to catch up. Suitable for high school students in Grades 10-12.

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