A particle starts from rest and accelerates uniformly for sometime it then stop by uniformly retarding. if the total time of journey is 8 second and retarding is 3 times of acceleration then the time for which it accelerates will be
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Heya user !!
Here's the answer you are looking for
See the v-t graph of this particle (see the attachment).
Let the acceleration is a, and it accelerates for time t.
Since, the retardation is 3 times acceleration,
retardation is (-3a) and time is (8-t) (as total time is t)
Let in time t, the particle accelerates to a velocity v.
☛ From the 1st equation of motion,
v = u + at
So, here during acceleration
v = 0 + at
v = at
And during retardation, as it comes to rest, final velocity is 0 and initial is v.
So,
0 = v + (-3a)(8-t)
v = 24a - 3at
☛ From both the equations of v, we get,
at = 24a - 3at
4at = 24a
4t = 24
t = 6
➡️ Therefore, the particle accelerates for 6sec.
★★ HOPE THAT HELPS ☺️ ★★
Here's the answer you are looking for
See the v-t graph of this particle (see the attachment).
Let the acceleration is a, and it accelerates for time t.
Since, the retardation is 3 times acceleration,
retardation is (-3a) and time is (8-t) (as total time is t)
Let in time t, the particle accelerates to a velocity v.
☛ From the 1st equation of motion,
v = u + at
So, here during acceleration
v = 0 + at
v = at
And during retardation, as it comes to rest, final velocity is 0 and initial is v.
So,
0 = v + (-3a)(8-t)
v = 24a - 3at
☛ From both the equations of v, we get,
at = 24a - 3at
4at = 24a
4t = 24
t = 6
➡️ Therefore, the particle accelerates for 6sec.
★★ HOPE THAT HELPS ☺️ ★★
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