i. What is maximum velocity attained by particle

ii. Time when particle will be at point of reversal

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : DEC 2015

FMAX = 10 N & aMAX = 10 m/s2

Observing graph till 10 seconds, we can write

$F = 2t ----(for\space t=0\space to\space t=10) $

Also, $a=2t (as\space m=1) $

Now, $dv/dt = a $

Hence, $v - u = \int a\space dt. ...(from\space t=0\space to\space t=10) \\ v - 10 = \int 2t\space 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 $ & $a= -10 m/s2 $

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.