Math Problem Statement

Two cars are travelling in opposite directions on a narrow north-south street. One is moving at 16 m/s [N] and the other at 12 m/s [S]. When the cars are 50 m apart, the two drivers apply their brakes simultaneously and slow down at the same rate. What is the magnitude of the minimum acceleration to avoid a collision?

Solution

To solve this, let's first outline the scenario:

  • Initial speeds: One car is traveling north at 16 m/s, and the other is traveling south at 12 m/s.
  • Initial distance: The cars are 50 meters apart.
  • Simultaneous braking: Both cars apply their brakes at the same time and decelerate at the same rate.

We want to find the minimum acceleration (deceleration in this case) required for the cars to avoid a collision.

Step-by-step approach:

  1. Relative velocity:
    Since the cars are moving toward each other, their relative velocity is the sum of their individual velocities: vrel=16m/s+12m/s=28m/sv_{\text{rel}} = 16 \, \text{m/s} + 12 \, \text{m/s} = 28 \, \text{m/s}

  2. Time to collision without braking: Without any deceleration, the time it would take for the cars to collide is given by: tcollision=distancerelative velocity=50m28m/s=1.7857secondst_{\text{collision}} = \frac{\text{distance}}{\text{relative velocity}} = \frac{50 \, \text{m}}{28 \, \text{m/s}} = 1.7857 \, \text{seconds}

  3. Deceleration needed to avoid collision: For the cars to avoid a collision, the relative velocity between them must reach zero. Let's denote the deceleration of each car as aa (same for both cars, since they brake simultaneously).

    The cars need to reduce their relative velocity from 28 m/s to 0 m/s in the time tcollisiont_{\text{collision}}. Using the equation for velocity under constant acceleration: vfinal=vinitial+atv_{\text{final}} = v_{\text{initial}} + a \cdot t where vfinal=0v_{\text{final}} = 0 m/s (since the final relative velocity must be zero), vinitial=28v_{\text{initial}} = 28 m/s, and t=1.7857t = 1.7857 s.

    Thus, the equation becomes: 0=28m/s+a1.7857s0 = 28 \, \text{m/s} + a \cdot 1.7857 \, \text{s} Solving for aa: a=28m/s1.7857s=15.7m/s2a = \frac{-28 \, \text{m/s}}{1.7857 \, \text{s}} = -15.7 \, \text{m/s}^2

    The negative sign indicates deceleration (slowing down). The magnitude of the required acceleration (deceleration) is: a=15.7m/s2|a| = 15.7 \, \text{m/s}^2

Final answer:

The magnitude of the minimum deceleration (acceleration) needed for both cars to avoid a collision is 15.7 m/s².


Do you need further clarification or a detailed explanation of any part of the solution?

Here are 5 questions related to this problem:

  1. How does relative velocity affect the time to collision?
  2. What are the assumptions made when calculating the time to collision?
  3. How can you calculate the deceleration needed for a different initial distance between the cars?
  4. What is the relationship between braking force and acceleration in this scenario?
  5. How would the problem change if the cars had different decelerations?

Tip: When calculating the time to collision, always consider the relative velocities of the objects involved, especially when they are moving toward each other.

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Math Problem Analysis

Mathematical Concepts

Relative Velocity
Kinematics
Acceleration
Motion Under Constant Acceleration

Formulas

Relative velocity: v_rel = v1 + v2
Time to collision: t_collision = distance / relative velocity
Equation of motion: v_final = v_initial + a * t

Theorems

Conservation of relative velocity
Uniform acceleration equation

Suitable Grade Level

Grades 9-12