The velocity of the particle is given by v = 4t^{2} + 7 cm/s. Find the velocity of the particle during the time interval t_{1} = 2s and t_{2} = 4s also find the average acceleration during the same interval of time.

### Asked by Topperlearning User | 17th Apr, 2014, 02:07: PM

Expert Answer:

v = 4t^{2} + 7

When t

_{1}= 2s, v_{1}= 4(2)^{2}+ 7 = 23 cm/sWhen t

_{2}= 4s, v_{2}= 4(4)^{2}+ 7 = 71 cm/sChange in velocity = v

_{2}– v_{1}= 71 – 23 = 48 cm/sAverage acceleration =

_{}### Answered by | 17th Apr, 2014, 04:07: PM

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