Question

A small projectile is fired vertically downward into a fluid medium with an initial velocity of 60 m/s. Due to the drag resistance of the fluid the projectile experiences a deceleration of a (-0.4) m/s2, where v is in m/s. Determine the projectile's velocity and position 4 s after it is fired.

Answer 1

Velocity after 4 seconds becomes
`58.4m/s`

Step-by-step explanation:

By definition of acceleration we have

`a=(dv)/(dt)`

It is given in our case that a =
`-0.4m/s^(2)`

Thus we have

`(dv)/(dt)=-0.4m/s^(2)\\ \\\\dv=-0.4dt\\\\Upon Integrating \\\\\int_(60)^{V_(f)}dv=\int_(0)^(4s)(-.4)dt`

Thus solving we get

`V_(f)-60=(-0.4)(4-0)\\ \\V_(f)=-1.6+60\\ \\V_(f)=58.4m/s`