A particle starts from rest and travels with constant acceleration 'a' . After a time of 't' it's acceleration direction gets reversed. Identify correct options
a) It's velocity when it returns back to initial point is √2 at
b) It's velocity when it returns back to initial point is 2at
c) Total distance travelled by particle when it reaches back to initial point is 2at^2
d) Time taken to come back to initial point (2+√2)t
Answers
Given : A particle starts from rest and travels with constant acceleration 'a' . After a time of 't' it's acceleration direction gets reversed.
To find : Correct options
Solution:
Particle starts from rest
u = 0
v = u + at = at
Distance covered = (1/2)at²
Now particle will moving in same direction until velocity is zero
0 = at - a(t₁)
=> t₁ = t
it moves for t more secs
Distance covered = at ² + (1/2)(-a)t² = (1/2)at²
Total Distance covered = (1/2)at² + (1/2)at² = at²
Come back to initial position Distance to be covered = -at²
Total Distance = |at²| + |-at²| = 2at²
Total distance travelled by particle when it reaches back to initial point is 2at²
using 2aS = V² = U²
2(-a)(-at²) = v² - (0)²
=> v² = 2a²t²
=> v = √2at
velocity when it returns back to initial point is √2 at
√2at = 0 + (a)t
=> t = √2t
Total time taken = t + t + √2t = (2+√2)t
Time taken to come back to initial point (2+√2)t
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Step-by-step explanation:
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