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The velocity function is v(t)= -t^2+5t-6 for a particle moving along a line. Find the distance traveled.
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I already got the displacement to be -67.5 I just need the distance traveled
differentiate with respect to t
can you work it out?
\[\frac{dv}{dt}=\frac{d(-2t ^{2})}{dt}+5\frac{d(t)}{dt}-\frac{d(6)}{dt}\]\[x=-4t+5\]it your required.
so now what
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?
Can someone please help me?
Distance travelled = \(\large \int_{t_1}^{t_2}V(t)dt\)
damn.. i did opposite.. @anthonys it is acceleration. yes yes.. you must integrate to find distance.
first put v(t)=0 you will get the time when it stops then integrate v(t) and put the value of t you will get your answer
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@experimentX thnx for correction.
happens sometime ... happens to me all the time.
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