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Q.

A particle is moving in x-y plane. At time t=0, particle is at (1 m, 2 m) and has velocity (4i^+6j^)m/s, At t=4 s, particle reaches at (6 m,4 m) and has velocity (2i^+10j^)m/s. In the given time interval, find average acceleration?

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a

(-0.5i^+6j^)m/s2

b

(-0.9i^+j^)m/s2

c

(-0.5+j^)m/s2

d

(-0.5i^+j^)m/s2

answer is D.

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Detailed Solution

aav=Δvt=vf-vit =(2i^+10j^)-(4i^+6j^)4 =(-0.5i^+j^)m/s2
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