Ch1 Ex 1d Q6 A motorbike and a car are waiting … 

The Problem:

A motorbike and a car are waiting side by side at traffic lights. When the lights turn to green, the motorbike accelerates at 2 ½ m/s² up to a top speed of 20 m/s, and the car accelerates at 1 ½ m/s² up to a top speed of 30 m/s. Both then continue to move at constant speed.

(a) After what time will the motorbike and the car again be side by side?

(b) What is the greatest distance that the motorbike is in front of the car?

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Douglas Quadling Mechanics1 Exercise1D Q6

Douglas Quadling Mechanics1 Exercise1D Q6

Douglas Quadling Mechanics1 Exercise1D Q6

Third Method:

Problem Statement

A motorbike and a car are waiting side by side at traffic lights. When the lights turn green:

• Motorbike: Accelerates at 2.5 m/s² up to a top speed of 20 m/s.
• Car: Accelerates at 1.5 m/s² up to a top speed of 30 m/s.

Both then continue at constant speed.

Questions:

1. After what time will the motorbike and the car again be side by side?
2. What is the greatest distance that the motorbike is in front of the car?

Solution

Part (a): Time When Both Vehicles Are Side by Side Again

To find when the motorbike and car are side by side, we calculate the time when both have traveled the same distance.

Step 1: Time to Reach Top Speed

• Motorbike: Time = 20 / 2.5 = 8 seconds
• Car: Time = 30 / 1.5 = 20 seconds

Step 2: Distance Traveled During Acceleration

• Motorbike: Distance = 0.5 * 2.5 * (8)^2 = 80 meters
• Car: Distance = 0.5 * 1.5 * (20)^2 = 300 meters

Step 3: Distance Traveled at Constant Speed

Let t be the total time.

• Motorbike: Distance = 80 + 20 * (t – 8)
• Car: Distance = 300 + 30 * (t – 20)

Step 4: Set Distances Equal and Solve for t

80 + 20(t – 8) = 300 + 30(t – 20)

Solving this equation gives t = 22 seconds.

Answer to Part (a): The motorbike and car will be side by side again after 22 seconds.

Part (b): Greatest Distance the Motorbike is in Front of the Car

To find the maximum distance the motorbike leads the car, we analyze the time when the motorbike is ahead.

Step 1: Time Interval When Motorbike is Ahead

The motorbike is ahead from t = 0 to t = 22 seconds.

Step 2: Finding the Time of Maximum Lead

The maximum lead occurs when the motorbike is still accelerating, and the car is accelerating at a slower rate.

Using the distance functions:

• Motorbike: Distance = 1.25 * t^2 (for t ≤ 8 seconds)
• Car: Distance = 0.75 * t^2 (for t ≤ 20 seconds)

The difference in distance is:

D(t) = 1.25 * t^2 – 0.75 * t^2 = 0.5 * t^2

At t = 8 seconds, D(8) = 0.5 * 64 = 32 meters.

For 8 < t ≤ 20 seconds:

D(t) = 20t – 80 – 0.75t^2

The maximum occurs at t ≈ 13.33 seconds, where D(t) ≈ 53.33 meters.

Answer to Part (b): The greatest distance the motorbike is in front of the car is 53.33 meters.

Summary

• Part (a): The motorbike and car will be side by side again after 22 seconds.
• Part (b): The greatest distance the motorbike is in front of the car is 53.33 meters.