Of course! Let’s solve the problem step-by-step.
Given:
- The velocity of the particle at point [tex]\(P\)[/tex] is [tex]\(40 \, \text{m/s}\)[/tex].
- The particle comes to rest at point [tex]\(R\)[/tex], so the velocity at [tex]\(R\)[/tex] is [tex]\(0 \, \text{m/s}\)[/tex].
- The distance from point [tex]\(Q\)[/tex] to point [tex]\(R\)[/tex] is [tex]\(50 \, \text{m}\)[/tex].
1. Finding the deceleration:
Let’s denote the deceleration by [tex]\(a\)[/tex].
We can use the kinematic equation that relates velocity, acceleration, and distance:
[tex]\[v^2 = u^2 + 2as\][/tex]
Here:
- [tex]\(v\)[/tex] is the final velocity,
- [tex]\(u\)[/tex] is the initial velocity,
- [tex]\(a\)[/tex] is the acceleration (deceleration, in this case),
- [tex]\(s\)[/tex] is the displacement.
At point [tex]\(R\)[/tex]:
[tex]\[0 = (40)^2 + 2a(70)\][/tex]
(The distance [tex]\(PQ + QR = 20 \text{ m} + 50 \text{ m} = 70 \text{ m}\)[/tex])
[tex]\[0 = 1600 + 140a\][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[140a = -1600\][/tex]
[tex]\[a = \frac{-1600}{140} = -\frac{1600}{140} \approx -11.43 \, \text{m/s}^2\][/tex]
Thus, the deceleration is [tex]\(11.43 \, \text{m/s}^2\)[/tex].
2. Finding the distance [tex]\(PQ\)[/tex]:
Using the same kinematic equation:
[tex]\[v^2 = u^2 + 2as\][/tex]
- [tex]\(v = 0 \, \text{m/s}\)[/tex] (velocity at [tex]\(Q\)[/tex]),
- [tex]\(u = 40 \, \text{m/s}\)[/tex] (velocity at [tex]\(P\)[/tex]),
- [tex]\(a = -11.43 \, \text{m/s}^2\)[/tex],
- [tex]\(s = PQ\)[/tex] (distance we want to find).
Rearranging the equation:
[tex]\[0 = (40)^2 + 2(-11.43)s\][/tex]
[tex]\[0 = 1600 - 22.86s\][/tex]
[tex]\[22.86s = 1600\][/tex]
[tex]\[s = \frac{1600}{22.86} \approx 70 \, \text{m}\][/tex]
Thus, the distance [tex]\(PQ\)[/tex] is approximately [tex]\(70 \, \text{m}\)[/tex].
3. Finding the time taken to cover [tex]\(PQ\)[/tex]:
We can use another kinematic equation to find the time:
[tex]\[v = u + at\][/tex]
Where:
- [tex]\(v = 0 \, \text{m/s}\)[/tex] (velocity at [tex]\(Q\)[/tex]),
- [tex]\(u = 40 \, \text{m/s}\)[/tex] (velocity at [tex]\(P\)[/tex]),
- [tex]\(a = -11.43 \, \text{m/s}^2\)[/tex],
- [tex]\(t\)[/tex] is the time taken to travel from [tex]\(P\)[/tex] to [tex]\(Q\)[/tex].
Rearranging the equation:
[tex]\[0 = 40 + (-11.43)t\][/tex]
[tex]\[11.43t = 40\][/tex]
[tex]\[t = \frac{40}{11.43} \approx 3.5 \, \text{seconds}\][/tex]
Therefore, the time taken to cover [tex]\(PQ\)[/tex] is approximately [tex]\(3.5\)[/tex] seconds.
In summary:
a. The distance [tex]\(PQ\)[/tex] is approximately [tex]\(70\)[/tex] meters.
b. The time taken to cover [tex]\(PQ\)[/tex] is approximately [tex]\(3.5\)[/tex] seconds.
