A point starts moving in a straight line with a certain acceleration. At a time t after beginning of
motion the acceleration suddenly becomes retardation of the same value. The time in which
the point returns to the initial point is
(A) 12t
(B) (2 + 2)
(C)
I
(D)
Cannot be predicted unless acceleration is given
Answers
Answer:
Explanation:
Using the second equation of motion- s = ut + 1/2 at² ,
where u = 0 and time = t1
When the acceleration changes, the distance travelled will be:
= s = 1/2 a t1²
When the acceleration changes to -a, then the particle continues in the same direction until the velocity will become zero. There after the particle changes the direction, starts to accelerate and passes over the point of start.
Using first equation of motion - v = u + a t
where v = 0 and acceleration = -a
= 0 = a t1 - a t
= t = t1 sincw, it takes t1 more time to stop and reverse direction.
Hence, the distance traveled/displacement in this time:
= s = u t + 1/2 at²
= s = a t1 ×t1 - 1/2 at1² = 1/2 a t1²
Total displacement from the initial point : 1/2 a t1² + 1/2 a t1² = a t1²
In negative direction - s = u t + 1/2 a t²
= -a t1² = 0 - 1/2 a t²
= t = √2 t1
The total time T from initial point forward till back to initial point is
T = 2 t1 + √2 t1
= (2 + √2) t1.
Explanation:
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