A man is running at 5 m/s. From a common starting point, a car is also moving at 5 kph and started to accelerate after 10 seconds to 50 kph within 5 seconds. If the man started ahead by 3 minutes, how long will it take before the car catches up? What is the displacement of the car and the man when they meet?




1. How many seconds will it take for a car to catch up with the man?
2. What are the displacement of the man and the car?

Let's break this problem down step by step. We'll focus on three phases:

1. **The man starts ahead by 3 minutes (180 seconds) at a constant speed of 5 m/s.**
2. **The car starts at 5 km/h (which is roughly 1.39 m/s), then accelerates to 50 km/h (which is roughly 13.89 m/s) after 10 seconds.**

### Phase 1: Displacement of the man
Since the man is running at a constant speed of 5 m/s and he has a 3-minute head start, we calculate his displacement after 3 minutes (180 seconds).

\[
\text{Displacement of the man after 180 seconds} = 5 \, \text{m/s} \times 180 \, \text{s} = 900 \, \text{meters}
\]

### Phase 2: Motion of the car

1. **For the first 10 seconds,** the car is moving at a constant speed of 5 km/h or 1.39 m/s. The displacement in this period is:

\[
\text{Displacement of the car in the first 10 seconds} = 1.39 \, \text{m/s} \times 10 \, \text{s} = 13.9 \, \text{meters}
\]

2. **From 10 to 15 seconds,** the car accelerates uniformly from 5 km/h to 50 km/h. The final velocity \(v_f\) is 13.89 m/s, and the initial velocity \(v_i\) is 1.39 m/s. The acceleration \(a\) is given by:

\[
a = \frac{v_f - v_i}{t} = \frac{13.89 \, \text{m/s} - 1.39 \, \text{m/s}}{5 \, \text{s}} = 2.5 \, \text{m/s}^2
\]

The displacement during this period can be found using the formula:

\[
\text{Displacement} = v_i t + \frac{1}{2} a t^2
\]

\[
\text{Displacement from 10 to 15 seconds} = 1.39 \, \text{m/s} \times 5 \, \text{s} + \frac{1}{2} \times 2.5 \, \text{m/s}^2 \times (5 \, \text{s})^2 = 6.95 \, \text{meters} + 31.25 \, \text{meters} = 38.2 \, \text{meters}
\]

3. **After 15 seconds,** the car moves at a constant speed of 50 km/h or 13.89 m/s. Now, we need to find out when the car will catch up with the man.

### Phase 3: Time until the car catches up

At 15 seconds, the total displacement of the car is:

\[
\text{Total displacement of the car after 15 seconds} = 13.9 \, \text{meters} + 38.2 \, \text{meters} = 52.1 \, \text{meters}
\]

At this point, the man is still 900 meters ahead. The car now moves at a constant speed of 13.89 m/s, and the man is moving at 5 m/s. The relative speed between the car and the man is:

\[
\text{Relative speed} = 13.89 \, \text{m/s} - 5 \, \text{m/s} = 8.89 \, \text{m/s}
\]

The time it will take for the car to catch up is:

\[
t = \frac{\text{Distance between car and man}}{\text{Relative speed}} = \frac{900 \, \text{meters} - 52.1 \, \text{meters}}{8.89 \, \text{m/s}} = \frac{847.9 \, \text{meters}}{8.89 \, \text{m/s}} \approx 95.36 \, \text{seconds}
\]

### Final Result:
1. **Time to catch up:** The car will catch up with the man approximately 95.36 seconds after the car reaches its constant speed (i.e., after the first 15 seconds).

2. **Displacement when they meet:**
   - The displacement of the man after 180 + 15 + 95.36 seconds is:
   
\[
\text{Displacement of the man} = 5 \, \text{m/s} \times (180 + 15 + 95.36) \approx 5 \, \text{m/s} \times 290.36 \, \text{s} = 1451.8 \, \text{meters}
\]

   - The displacement of the car is the same as the man’s displacement when they meet, which is approximately **1451.8 meters**.

### Summary:
- **Time until the car catches up:** 110.36 seconds (15 seconds + 95.36 seconds).
- **Displacement of the man and car when they meet:** 1451.8 meters.

Would you like further clarification or more details? Here are some related questions:

1. How did we convert speeds from km/h to m/s in this scenario?
2. What role did uniform acceleration play in the car’s catching up?
3. How do you calculate displacement for an accelerating object?
4. Can the car catch up faster if it accelerates sooner?
5. What would happen if the man’s speed increased during the race?

**Tip:** Always convert speeds to consistent units (e.g., meters per second) before solving.

A man is running at 5 m/s. From a common starting point, a car is also moving at 5 kph and started to accelerate after 10 seconds to 50 kph within 5 seconds. If the man started ahead by 3 minutes, how long will it take before the car catches up? What is the displacement of the car and the man when they meet?

Solve this kinematics problem involving a man running at 5 m/s and a car accelerating from 5 kph to 50 kph. Learn how to calculate displacement, acceleration, and the time it takes for the car to catch up after the man’s 3-minute head start. Suitable for students in Grades 10-12.

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