Math Problem Statement

solve this

A cyclist starts from rest and accelerates uniformly down a slope. After covering a distance of x meters, the cyclist reaches a speed of 6 m/s. If the acceleration of the cyclist is 1.8 m/s², calculate the value of x.

Solution

This problem involves the use of the kinematic equation for uniformly accelerated motion. The equation that relates the distance traveled xx, initial velocity uu, final velocity vv, and acceleration aa is:

v2=u2+2axv^2 = u^2 + 2ax

Given:

  • Initial velocity, u=0u = 0 m/s (since the cyclist starts from rest)
  • Final velocity, v=6v = 6 m/s
  • Acceleration, a=1.8a = 1.8 m/s2^2

We need to find the distance xx traveled by the cyclist.

Applying the kinematic equation:

v2=u2+2axv^2 = u^2 + 2ax

Substituting the given values:

(6)2=(0)2+2(1.8)x(6)^2 = (0)^2 + 2(1.8)x

36=3.6x36 = 3.6x

Solving for xx:

x=363.6=10metersx = \frac{36}{3.6} = 10 \, \text{meters}

Conclusion:

The correct answer is Option D: 10 m.

Would you like more details or have any questions? Here are some related questions to consider:

  1. What would the distance xx be if the final velocity was 8 m/s instead of 6 m/s?
  2. How would the distance change if the acceleration was reduced to 1 m/s²?
  3. What would happen to the distance if the initial velocity was not zero?
  4. Can you derive the kinematic equation used in this problem?
  5. How is the motion affected if the cyclist experiences a deceleration instead of acceleration?

Tip: In uniformly accelerated motion, the distance covered is directly proportional to the square of the final velocity when starting from rest.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Uniform Acceleration

Formulas

v^2 = u^2 + 2ax

Theorems

Kinematic equations of motion

Suitable Grade Level

Grade 9-11