A particle is moving with initial velocity 40m/s along positive X-axis Acceleration is -10m/s2 It starts from - Physics - Motion In A Straight Line

Question

A particle is moving with initial velocity 40m/s along positive
X-axis Acceleration is -10m/s2 It starts from x =10 m Find out
maximum co-ordinate of particle in positive direction (a) 90 m (b)
0 m (c) 120 m (d) 30 m

Jyoti Pant · Accepted Answer

Here,u=40m/s,For the maximum coordinate,v=0 and a=- $10 m / s^{2}$

Now,by using

$v^{2} = u^{2} + 2 a s 0 = 40^{2} + 2 \times (- 10) \times s 0 = 1600 - 20 s s = 1600 / 20 = 80 m$

Since the particle starts from x=10m,so the maximum distance or coordinate from the origin=80+10=90m

A particle is moving with initial velocity 40m/s along positive X-axis. Acceleration is -10m/s2. It starts from x =10 m. Find out maximum co-ordinate of particle in positive direction. (a) 90 m (b) 0 m (c) 120 m (d) 30 m.