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CBSE Class 11-science Answered

A particle travels in a straight line, such that for a short time 2s≤t≤6s, its motion is described by v=(4/a) m/s, where a is in m/s 2 . If v =8m/s when t = 2s, determine the particle's acceleration when t = 4s.

Asked by varmaakshay2020 | 13 Jun, 2022, 11:43: AM
Expert Answer
Velocity v = ( 4/a ) m /s  ....................... (1)
 
where a = acceleration in m/s2
 
Hence we get , v × a =  v × (dv/dt) = 4 
 
v × dv = 4 dt
 
By integrating both sides of above expression, we get
 
(1/2) v2 = 4 t + C  ........................(2)
 
where C is integration constant
 
If we apply initial condition , at t = 2 , v = 8 to above expression, we get C = 24
 
hence ,  eqn.(2) is rewritten as  v2 = 8t + 48  .................... (3)
 
velocity at t = 4 s is calculated from eqn.(3) as ,  v = 4√5 m/s
 
Acceleration at t = 4 is calculated from eqn.91) as ,  a = 4 / ( 4√5 ) = ( 1 / √5 ) m/s2
Answered by Thiyagarajan K | 13 Jun, 2022, 12:34: PM
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