A particle starts from rest and moves with a constant acceleration of ...
Solution:
Given:
Initial velocity (u) = 0
First acceleration (a1) = A1
Second acceleration (a2) = -A2 (deceleration)
Time for which particle is in motion (t) = T1
Formula:
Distance = ut + (1/2)at^2 (when acceleration is constant)
Distance = (1/2)(u+v)t (when acceleration is not constant)
Calculation:
Phase 1: Constant acceleration (A1)
Distance covered in time T1/2 = (1/2)*A1*(T1/2)^2 = (1/8)*A1*T1^2
Velocity after T1/2 time = u + A1*(T1/2) = (1/2)*A1*T1
Phase 2: Deceleration (-A2)
Let the time taken to come to rest be T2
Final velocity = u + a*t = 0
=> -A2*T2 = -(1/2)*A1*T1
=> T2 = (1/2)*(A1/A2)*T1
The distance covered in time T2 = (1/2)*(-A2)*(T2)^2 = (1/8)*(A1/A2)^2*T1^2
The total distance travelled by the particle is the sum of distances covered in phase 1 and phase 2.
Total distance = (1/8)*A1*T1^2 + (1/8)*(A1/A2)^2*T1^2
= (1/8)*T1^2*(A1 + A1^2/A2^2)
Therefore, the total distance covered by the particle is (1/8)*T1^2*(A1 + A1^2/A2^2).
A particle starts from rest and moves with a constant acceleration of ...
Total distance = V^2+u^2/2(A1+A2)m .....(In first case the distance is x=v^2/2A1,,In second case the distance is u^2/2A2 due to constant acceleration ).(And T1=0). I hope it help !!....