Math Problem Statement

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.5 m/s , and the distance between them is 90.0 m . After t1 = 3.00 s , the motorcycle starts to accelerate at a rate of 4.00 m/s2 . The motorcycle catches up with the car at some time t2 How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2−t1 .

Solution

We need to find the time it takes for the motorcycle to catch up with the car after the motorcycle starts accelerating.

Given:

  • Initial speed of both the car and the motorcycle: v0=19.5m/sv_0 = 19.5 \, \text{m/s}
  • Initial distance between the car and the motorcycle: d0=90.0md_0 = 90.0 \, \text{m}
  • The motorcycle starts accelerating after t1=3.00st_1 = 3.00 \, \text{s}.
  • Acceleration of the motorcycle: a=4.00m/s2a = 4.00 \, \text{m/s}^2

To find:

  • The time Δt=t2t1\Delta t = t_2 - t_1, where t2t_2 is the time when the motorcycle catches up with the car after the motorcycle starts accelerating.

Step-by-step solution:

  1. Position of the car: The car is moving at a constant speed, so its position at any time tt after t1t_1 can be described by: xcar(t)=v0tx_{\text{car}}(t) = v_0 \cdot t

  2. Position of the motorcycle: After the motorcycle starts accelerating at t=t1t = t_1, its position at any time tt after it starts accelerating is: xmotorcycle(t)=v0t+12a(tt1)2x_{\text{motorcycle}}(t) = v_0 \cdot t + \frac{1}{2} a \cdot (t - t_1)^2 where a=4.00m/s2a = 4.00 \, \text{m/s}^2 is the acceleration.

  3. Catching up condition: The motorcycle catches up to the car when their positions are the same: xmotorcycle(t2)=xcar(t2)x_{\text{motorcycle}}(t_2) = x_{\text{car}}(t_2) Substituting the expressions for xcar(t)x_{\text{car}}(t) and xmotorcycle(t)x_{\text{motorcycle}}(t) into this equation gives: v0t2+12a(t2t1)2=v0t2v_0 \cdot t_2 + \frac{1}{2} a \cdot (t_2 - t_1)^2 = v_0 \cdot t_2

  4. Solve for t2t1t_2 - t_1: Since v0t2v_0 \cdot t_2 cancels out on both sides, we're left with: 12a(t2t1)2=d0\frac{1}{2} a \cdot (t_2 - t_1)^2 = d_0 Substituting the given values a=4.00m/s2a = 4.00 \, \text{m/s}^2 and d0=90.0md_0 = 90.0 \, \text{m}, we have: 124.00(t2t1)2=90.0\frac{1}{2} \cdot 4.00 \cdot (t_2 - t_1)^2 = 90.0 Simplifying: 2.00(t2t1)2=90.02.00 \cdot (t_2 - t_1)^2 = 90.0 (t2t1)2=90.02.00(t_2 - t_1)^2 = \frac{90.0}{2.00} (t2t1)2=45.0(t_2 - t_1)^2 = 45.0 Taking the square root: t2t1=45.06.71secondst_2 - t_1 = \sqrt{45.0} \approx 6.71 \, \text{seconds}

Final answer:

The time it takes for the motorcycle to catch up with the car after it starts accelerating is approximately 6.71seconds6.71 \, \text{seconds}.

Would you like more details or have any questions?

Here are some related questions you could explore:

  1. What is the distance traveled by the motorcycle while catching up?
  2. How fast is the motorcycle traveling when it catches up?
  3. What if the motorcycle accelerated at a different rate?
  4. How much total time does it take for the motorcycle to catch up, including the first 3 seconds?
  5. What if the initial distance was different?

Tip: When analyzing motion problems, breaking them into phases (constant speed, acceleration, etc.) makes solving much easier!

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Math Problem Analysis

Mathematical Concepts

Kinematics
Quadratic Equations
Relative Motion

Formulas

Position of the car: x_car(t) = v_0 * t
Position of the motorcycle: x_motorcycle(t) = v_0 * t + 1/2 * a * (t - t_1)^2
Catching up condition: x_motorcycle(t2) = x_car(t2)

Theorems

Equations of motion under constant acceleration

Suitable Grade Level

Grades 10-12