Math Problem Statement

Two particles P and Q are moving along the same straight horizontal line with constant accelerations 2 m/s² and 3.6 m/s² respectively. At time t = 0, P passes through point A with a speed of 4 m/s. One second later, Q passes through A with a speed of 3 m/s, moving in the same direction as P. Write down expressions for the displacements of P and Q from A, in terms of t, where t seconds is the time after P has passed through A.

Solution

The problem you've uploaded describes two particles, PP and QQ, moving along the same horizontal line with constant accelerations. The acceleration of PP is 2m/s22 \, \text{m/s}^2, while QQ has an acceleration of 3.6m/s23.6 \, \text{m/s}^2. The motion begins with:

  • At time t=0t = 0, PP passes through point AA with a speed of 4m/s4 \, \text{m/s}.
  • One second later, QQ passes through point AA with a speed of 3m/s3 \, \text{m/s}, and both particles are moving in the same direction.

Part (a) asks for expressions of displacement:

You are required to write down expressions for the displacements of PP and QQ from AA, in terms of tt, where tt represents the time after PP has passed through AA.

Displacement of particle PP:

For particle PP, the displacement after time tt can be given by the kinematic equation: sP=uPt+12aPt2s_P = u_P t + \frac{1}{2} a_P t^2 where:

  • uP=4m/su_P = 4 \, \text{m/s} (initial velocity of PP),
  • aP=2m/s2a_P = 2 \, \text{m/s}^2 (acceleration of PP),
  • tt is the time after PP has passed through AA.

Thus, the displacement of PP is: sP=4t+12×2×t2=4t+t2s_P = 4t + \frac{1}{2} \times 2 \times t^2 = 4t + t^2

Displacement of particle QQ:

For particle QQ, since it passes through AA one second later, the time elapsed for QQ is t1t - 1 (i.e., t1t - 1 seconds have passed for QQ when PP has traveled for tt seconds). The displacement of QQ can also be calculated using the kinematic equation: sQ=uQ(t1)+12aQ(t1)2s_Q = u_Q (t - 1) + \frac{1}{2} a_Q (t - 1)^2 where:

  • uQ=3m/su_Q = 3 \, \text{m/s} (initial velocity of QQ),
  • aQ=3.6m/s2a_Q = 3.6 \, \text{m/s}^2 (acceleration of QQ).

Thus, the displacement of QQ is: sQ=3(t1)+12×3.6×(t1)2s_Q = 3(t - 1) + \frac{1}{2} \times 3.6 \times (t - 1)^2 Simplifying: sQ=3(t1)+1.8(t1)2s_Q = 3(t - 1) + 1.8(t - 1)^2

Summary:

The expressions for the displacements are:

  • sP=4t+t2s_P = 4t + t^2
  • sQ=3(t1)+1.8(t1)2s_Q = 3(t - 1) + 1.8(t - 1)^2

Would you like further clarification or the solution to part (b) of the question?

Relative questions:

  1. What would be the displacement of PP after 5 seconds?
  2. How do the displacements of PP and QQ compare at t=2t = 2 seconds?
  3. Can you find the time at which both particles PP and QQ are at the same position?
  4. How would the problem change if the initial speed of QQ were greater than PP?
  5. What is the physical significance of the acceleration for each particle in this problem?

Tip:

When solving problems with relative motion, always carefully account for the time difference between the movements of objects.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Displacement
Time-dependent motion
Algebra

Formulas

Displacement formula: s = ut + 1/2 at²
Displacement for P: s_P = 4t + t²
Displacement for Q: s_Q = 3(t - 1) + 1.8(t - 1)²

Theorems

Equations of motion under constant acceleration

Suitable Grade Level

Grades 10-12