Question

The position of a particle as a function of time t is given by
x=5t³-12t
then acceleration of the particle
Options
1. Remains constant
2. Increases with time
3. Decreases with time
4. First increases and then decreases with time

please answer fast which option is correct please ​

Answer 1

Answer:

The displacement is given as x=5t

2

+4t+3

⇒x−3=5t

2

+4t

assuming x−3=s,

s=4t+

2

1

10t

2

comparing this with s=ut+

2

1

at

2

we get u=4m/s and a=10m/s

2

so velocity at t=2sec is v=u+at=4+10×2=24m/s