Question

# A particle starting from rest undergoes a rectilinear motion with acceleration a. The variation of a with time ( t ) is shown below. The maximum velocity attained by the particle during its motion is

Solution

From the diagram we can expless equation as

[

a=10-left(frac{10}{12}right) t

]

where at ( t=0 ) as ( 10 mathrm{m} / mathrm{s}^{2}, t=12 ) a ( =0 mathrm{m} / mathrm{s}^{2} )

by integration we get

[

begin{array}{l}

int a=int 10-left(frac{0}{12}right) t

V=10 t-left(frac{10}{12}right) frac{t^{2}}{2}

V=10 t-frac{5 t^{2}}{12}

end{array}

]

fol ( operatorname{Vmax} frac{d v}{d t}=0 )

[

begin{array}{l}

Rightarrow frac{d V}{d t}=10-frac{5}{12} times 2 times t

0=10-frac{5 t}{6} Rightarrow frac{1}{6}=100-1 t=12 mathrm{sec}

end{array}

]

Volocity at 12 sec

[

V=(10 times 12)-frac{5}{12} times 12 times 12

]

[

V=120-60=60 mathrm{m} / mathrm{s}

]

Vmax

[

text { Areaculdergraph }=frac{1}{2} times 12 times 10=60 mathrm{m} / mathrm{s}

]