Question

난
( frac{1}{(12)} )
( a-0=(t-12)left(frac{10-0}{0-12}right) )
( a=(t-12)left(frac{-5}{6}right) )
( 26 a=-5 t+60 )
( a=frac{d v}{d x}=frac{-5 t}{6}+10 )
94035
velowty ( mathcal{W} )
( omega ln n=0 )
( frac{5 t}{6}=frac{r_{0}}{t=12} )
Why ratuy
[
frac{v=-5 t^{2}}{12}+10 t+c
]
( t=0 quad v=0 )
( v=frac{-5}{x} times 12 times 14+10 times 12 )
( begin{array}{rl}2 & -60 V_{max } 2 & 60end{array} )

# 16. 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 10 a (m/s) 12 t's

Solution