The velocity-time graph of a particle moving along a straight line is shown. The rate of acceleration and deceleration is constant and it is equal to 5 metre per second square if the average velocity during the motion is 20 metre per second then the value of t is
(question no. 19)
Answers
Answer:
t = 5 sec
Explanation:
The velocity-time graph of a particle moving along a straight line is shown. The rate of acceleration and deceleration is constant and it is equal to 5 metre per second square if the average velocity during the motion is 20 metre per second then the value of t is
initial Velocity at t = 0 is 0 m/s
a = 5 m/s ²
using V = U + at
velocity after t = 0 + 5t = 5t m/s
after some time a = -5 m/s²
Time taken to Become 0 again
0 = 5t - 5t
=> time = t
time with constant Speed = 25 - 2t
Distance Covered in first T sec
S = (1/2)5t² = 5t²/2
Distance covered in 25 - 2t = 5t( 25 - 2t)
= 125t - 10t²
Distance Covered in Last T sec
= 5t*t - (1/2)5t² = 5t²/2
Total Distance = 5t²/2 + 125t - 10t² + 5t²/2 = 125t - 5t²
time = 25 sec
Average Speed = 20 m/s
20*25 = 500
=> 125t - 5t² = 500
=> t² - 25t = -100
=> t² - 25t + 100 = 0
=> t² - 5t - 20t + 100 = 0
=> (t -5)(t - 20) = 0
=> t = 5 or 20 sec
As per graph first value of t is asked
=> t = 5 sec
Answer:
t=5 hope this answer may help you ..
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