a particle moving with uniform acceleration has velocity, 6 m/s at a distance of 5 m from the initial position. after moving another 7 m, the velocity becomes 8 m/s. find the initial velocity and acceleration of the particle
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37
Let acceleration be a m/s^2 and initial velocity be u m/s.
Now s(distance) in first part is 5 m and final velocity is 6m/s
therefore, using- v^2-u^2=2as we get,
6^2-u^2=2*a*5
=> 36-u^2=10a --------- eq.1
now in second part of journey...6m/s becomes initial velocity, 8m/s becomes final velocity and 7m is s
again applying the same equation...we get,
8^2 - 6^2=2*a*7
=> 64-36=14a
=>28=14a
=>a=2m/s^2
NOW SUBSTITUTING VALUE OF a IN EQ. 1, WE GET-
36-u^2=10*2
=>36-u^2=20
=>36-20=u^2
=>16=u^2
=>u=4m/s
Therefore, we get- acceleration = 2m/s^2 and initial velocity= 4m/s
Now s(distance) in first part is 5 m and final velocity is 6m/s
therefore, using- v^2-u^2=2as we get,
6^2-u^2=2*a*5
=> 36-u^2=10a --------- eq.1
now in second part of journey...6m/s becomes initial velocity, 8m/s becomes final velocity and 7m is s
again applying the same equation...we get,
8^2 - 6^2=2*a*7
=> 64-36=14a
=>28=14a
=>a=2m/s^2
NOW SUBSTITUTING VALUE OF a IN EQ. 1, WE GET-
36-u^2=10*2
=>36-u^2=20
=>36-20=u^2
=>16=u^2
=>u=4m/s
Therefore, we get- acceleration = 2m/s^2 and initial velocity= 4m/s
Answered by
0
Answer: a=2 ms2
and u = 4ms
Explanation:
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