A particle is travelling in the positive x-axis direction with uniform acceleration for 6T seconds. It has an average velocity of 10 m/s for first T seconds. It has an average velocity of 40 m/s for the last 3T seconds. What is the average velocity for the middle 2T seconds?
Answers
Explanation:
Here we are given a = 6(t- 1) Here, s = +x, v = +v. To find velocity-time relation using a = dv/dt, we have dtdv = 6 ( t - 1 )...(i)When velocity changes from v0 to v during time t, integrating both sides with respect to time, we have∫v0vdv=∫0t6(t−1)dt⇒v−v0=3t2−6t Substituting v0 = 2 m/s, we have v = dtdx=3t2−6t+2 (ii)dx = (3t2−6t+2)dt.(ii) find position-time relation, again integrating (ii) both sides, we have ∫0xdx=∫0t(3t2−6t+2)dt This gives x=t3−3t2 + 2t = t (t - 1)(t - 2) iii)The particle passes through the origin where x = 0 from (iii), we have t = 0, 1s and 2s. That means the particle crosses the origin twice at t = 1 s and t = 2 s. After t = 2 s, x is positive. Hence, its displacement points in positive x-direction and goes on increasing its magnitude.The particle will come at rest when v = 0. From (ii), we have t = (1−31)s (=t1, say and t = (1+31)s (=t2, say)Hence, the particle at t =t1 and t2. That signifies to and fro motion of the particle during first two seconds as it starts retracing its path at t = t1 and t =
Answer:
21.5 m/s
Hint:
Acceleration is same throughout. so make 2 formulas for acceleration where T and average velocity are the only variables.
On solving, you will find that T cancels out and you get the value of average velocity as 21.5 m/s
( use your brains now ! )
hope it helps :)