i. What is maximum velocity attained by particle

ii. Time when particle will be at point of reversal

$$FMAX = 10 N \space\space \& \space\space aMAX = 10 m/s2 $$

Observing graph till 10 seconds, we can write

$F = 2t \text {__(for t=0 to t=10) } \\ \text {Also }, a=2t (as m=1) $

Now, $dv/dt = a $

Hence, $v - u = \int a dt. $...(from t=0 to t=10)

$$v - 10 = \int 2t dt $$

$v = [t2]_0^{10} + 10 \\ v = 110 m/s $

As after 10 sec force is changing its direction, velocity at 10 seconds will be maximum velocity therefore, $VMAX = 110 m/s $

Now, after 10 seconds, $F= -10N \space \space \&\space\space a= -10 m/s^2 $

At point of reversal velocity of particle will be 0

$V= u + at \\ 0 = 110 - 10(t) \\ t = 11 sec $

Maximum velocity attained by particle is 110 m/s & Time when particle will be at point of reversal is 11 sec.