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

Answer:

Velocity after 4 seconds becomes $58.4m/s$

Explanation:

By definition of acceleration we have

$a=\frac{dv}{dt}$

It is given in our case that a = $-0.4m/s^{2}$

Thus we have

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

Thus solving we get

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