A particle starts from position x = 0 at time t = 0
with velocity v = (t + 4t) along the x-axis. The
magnitude of average acceleration during the time
interval t = 1 s to t = 2 s is (where v is in m/s and
t in s)
6 = 7
(1) 8 ms-2
2) 7 ms2
(3) 6 ms-2
(4) Zero
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Answer is zero
when we differentiate this equuation w.r.t to time we get
dv/dt=(t+4t)/dt
da=1+4
as we can see there is no time term here.so it is independent of time
therefore answer is 0
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