The displacement y of body varies with time t as ÷ y=-2/3t2+16t+2 how long does the body take to come at rest
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Answered by
180
y=2/3t^2+16t+2
So,dy/dt=4/3t+16
So, time taken by object to come to rest
=> dy/dt=v=4/3t+16
=>4/3t+16=0 (as object is at rest)
=>t=(-12) sec
But time cant be negative so object will never come to rest..
PLZ MARK IT AS BRAINLIEST..
So,dy/dt=4/3t+16
So, time taken by object to come to rest
=> dy/dt=v=4/3t+16
=>4/3t+16=0 (as object is at rest)
=>t=(-12) sec
But time cant be negative so object will never come to rest..
PLZ MARK IT AS BRAINLIEST..
RaunakRaj:
plz mark it as brainliest
Hello @RaunakRaj , you missed the minus sign. Please correct your answer.
now i cant edit it.. so sry as the minus sign was mixed with = sign.
I know it sometimes happens. Even I can't edit my answers sometimes. But don't worry, your magnitude of final answer is correct. Just the sign is reversed
ok
i know that as i am also studying physics
Yeah,
Answered by
84
y = (-2/3) t^2 + 16 t + 2
So, velocity is
v = dy/dt = (-4/3) t + 16
The body comes at rest. So velocity must become zero.
So, v = (-4/3) t + 16 = 0
So, t = 12 s
Thus, the body will come to rest at t=12 seconds
So, velocity is
v = dy/dt = (-4/3) t + 16
The body comes at rest. So velocity must become zero.
So, v = (-4/3) t + 16 = 0
So, t = 12 s
Thus, the body will come to rest at t=12 seconds
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