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The speed(v) of a particle moving along a straight line is given by `v=(t^(2)+3t-4` where v is in m/s and t in seconds. Find time t at which the particle will momentarily come to rest. |
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Answer» When particle comes to rest, v=0. So `t^(2)+3t-4=0rArr t=(-3+-sqrt(9-4(1)(-4)))/(2(1))rArr t=1 "or" -4` Neglect negative value of t, Hence t=1 s |
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