Math Problem Statement

Two vehicles, K and L, are moving toward each other with speeds of 20 m/s and 40 m/s, respectively. They are initially 400 meters apart and both begin to decelerate at the same rate. What is the magnitude of the deceleration (in m/s²) such that both vehicles stop at the same point?

Solution

The problem in the image asks for the deceleration (negative acceleration) magnitude required for two vehicles, K and L, moving toward each other at 20 m/s and 40 m/s, respectively, to stop at the same position after initially being 400 meters apart. Both vehicles are said to decelerate with the same magnitude of acceleration.

Given dataThe problem in the image asks for the deceleration (negative acceleration) magnitude required for two vehicles, K and L, moving toward each other at 20 m/s and 40 m/s, respectively, to stop at the same position after initially being 400 meters apart. Both vehicles are said to decelerate with the same magnitude of acceleration.

Given

  • Speed of vehicle K: vK=20m/sv_K = 20 \, \text{m/s}
  • Speed of vehicle L: vL=40m/sv_L = 40 \, \text{m/s}
  • Distance between K and L: d=400md = 400 \, \text{m}
  • Both vehicles decelerate at the same magnitude aa, and we need to find this acceleration.

Solution Strategy:

We can use the equations of motion for uniformly accelerated motion. The stopping distance dd for an object with an initial velocity v0v_0 and a constant acceleration aa is given by:

d=v022ad = \frac{v_0^2}{2a}

Let the stopping distances of vehicles K and L be dKd_K and dLd_L, respectively. Since the total distance between them is 400 m, we have:

dK+dL=400md_K + d_L = 400 \, \text{m}

For vehicle K: dK=vK22a=(20)22a=4002a=200ad_K = \frac{v_K^2}{2a} = \frac{(20)^2}{2a} = \frac{400}{2a} = \frac{200}{a}

For vehicle L: dL=vL22a=(40)22a=16002a=800ad_L = \frac{v_L^2}{2a} = \frac{(40)^2}{2a} = \frac{1600}{2a} = \frac{800}{a}

Now, sum the distances:

200a+800a=400\frac{200}{a} + \frac{800}{a} = 400

1000a=400\frac{1000}{a} = 400

Solve for aa:

a=1000400=2.5m/s2a = \frac{1000}{400} = 2.5 \, \text{m/s}^2

Thus, the magnitude of the deceleration (acceleration) is 2.5m/s22.5 \, \text{m/s}^2, which corresponds to the answer:

B52\boxed{B} \, \frac{5}{2}

Would you like further details or clarification on any steps?

Here are 5 related questions to explore further:

  1. How would the answer change if one car had a different deceleration rate?
  2. What is the total time taken for both vehicles to stop?
  3. How would the problem change if the distance between the cars was 500 meters?
  4. Can this problem be solved graphically using velocity-time diagrams?
  5. What if one vehicle had a constant speed while the other decelerated?

Tip: In motion problems involving deceleration, using the equation d=v022ad = \frac{v_0^2}{2a} is a quick way to link initial speed, acceleration, and stopping distance.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Uniform Deceleration
Equations of Motion

Formulas

d = v_0² / (2a)
d_total = d_K + d_L

Theorems

Equation of motion under uniform acceleration

Suitable Grade Level

Grades 10-12