A particle starts from rest and travel a distance s with uniform accel...
Let a be the uniform acceleration.
v^2 - u^2 = 2 a S => S = v^2 / 2 a as u = 0.
final velocity = v = √(2 a S)
time taken to reach this velocity : t1 = (v - u ) / a = v / a = √(2S/a)
Now the particle moves with this velocity for distance 2 S.
so time taken t2 = 2 S / v = 2 S / √(2aS) = √(2 S / a)
Now the particle moves with a uniform deceleration - a3 for a distance 3 S.
time taken t3 to stop.
2 * a3 * 3 S = v^2 - u^2 here v = 0 and u = √(2 a S)
=> a3 = a / 3
=> t3 = - u /a3 = 3 √( 2 S / a)
total time duration = t = t1 + t2 + t3 = 5 * √(2 S / a)
total distance travelled : S + 2S + 3S =6 S
=> average speed = 3 / 5 * √(2 a S)
maximum speed = √(2 a S)
Ratio = 3/5
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A particle starts from rest and travel a distance s with uniform accel...
7a +2:a where a is the acceleration. but i'm not pretty sure it took too much calculation if you know plz tell
A particle starts from rest and travel a distance s with uniform accel...
The problem states that a particle starts from rest and undergoes three phases of motion: uniform acceleration, uniform speed, and uniform retardation. We need to find the ratio of its average velocity to maximum velocity.
Let's analyze each phase of motion separately:
Phase 1: Uniform acceleration
- The particle starts from rest, so its initial velocity (u) is 0.
- It travels a distance (s) with uniform acceleration.
- Let's assume the final velocity after this phase is (v1).
Using the equation of motion, we can relate the variables:
v1^2 = u^2 + 2as
Since u = 0, the equation simplifies to:
v1^2 = 2as
Phase 2: Uniform speed
- The particle travels a distance of 2s with uniform speed.
- The velocity remains constant throughout this phase.
- Let's assume the constant velocity is (v2).
Phase 3: Uniform retardation
- The particle travels a distance of 3s with uniform retardation.
- It finally comes to rest, so the final velocity (v3) is 0.
- Let's assume the initial velocity of this phase is (v2).
Using the equation of motion, we can relate the variables:
v3^2 = v2^2 - 2as
Since v3 = 0, the equation simplifies to:
0 = v2^2 - 2as
Combining the equations for phases 1 and 3, we can eliminate 'a' and solve for v2:
2as = v2^2
v2^2 - 2as = 0
Adding these equations, we get:
v1^2 + v2^2 = 2as + 0
v1^2 + v2^2 = 2as
Now, let's find the average velocity and maximum velocity:
Average velocity (Vavg) = Total distance / Total time
The total distance covered is s + 2s + 3s = 6s
To find the total time, we need to find the time taken for each phase. Assuming the acceleration is 'a' and the time taken for each phase is t1, t2, and t3 respectively:
t1 = v1 / a
t2 = 2s / v2
t3 = v2 / a
The total time (T) is:
T = t1 + t2 + t3
T = v1 / a + 2s / v2 + v2 / a
Maximum velocity (Vmax) occurs at the end of phase 1, which is v1.
The ratio of average velocity to maximum velocity is:
Vavg / Vmax = (6s / T) / v1
Vavg / Vmax = 6s / (v1 * T)
Now, substituting the expression for T:
Vavg / Vmax = 6s / (v1 * (v1 / a + 2s / v2 + v2 / a))
Simplifying this expression further would depend on the specific values of 'a', 's', 'v1', and 'v2'.
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