BSc Nursing GNM 2019 Solved Paper

Distance travelled by the particle is
x=40+12t−t3
We know that, velocity is rate of change of distance, i.e.
v‌‌=‌

dx

dt

∴‌‌v‌‌=‌

d

dt

(40+12t−t3)
‌‌=0+12−3t2
But final velocity, v=0
∴12−3t2=0
‌ or ‌t2=‌

12

3

=4
or
t=2s
Hence, distance travelled by the particle before coming to rest is given by
x‌‌=40+12(2)−(2)3
‌‌=40+24−8=64−8
‌‌=56m