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Acceleration-time graph of a particle moving in a straight line is as shown in figure. At time `t = 0,` velocity of the particle is zero. Find (a) average acceleration in a time interval from `t = 6 s` to `t = 12 s,` (b) velocity of the particle at `t = 14 s.` |
Answer» Correct Answer - A::B (a) `a_(av)=(v_f-v_i)/t=(v_12-v_6)/(12-6)` `=((-10)-(20))/6=-5m//s^2` (b)`Deltav=v_f-v_i`=net area of a-t graph `:. v_(14)-v_0=40+30+40-20=90` But `v_0=0` `:. v_(14)=90m//s` |
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