A soldier fires a warning shot straight up into the air with pistol. (Projectile Problem Differentation )

Question

A soldier fires a warning shot straight up into the air with pistol. The height (in feet) after t

seconds of the shot is given by

i. Find the velocity as a function of time.

ii. Find the acceleration as a function of time.

iii. Determine the position of the shot when the velocity is 0?

Answer 1

Velocity is the slope of the position curve, or, in other words, the derivative of the position function.

s'(t) = v(t) = -20/3t + 17/6

Acceleration is the slope of the velocity curve, or, in other words, the derivative of the velocity function (second derivative of the position function):

s''(t) = v'(t) = a(t) = -20/3

To find the position when velocity is zero, first plug in v = 0 into the velocity functio and find out the associated time, then plug that time into the position function to find position:

0 = -20/3t + 17/6

20/3t = 17/6

t = 17/40

s(17/40) = 1729/480 feet ≈ 3.602 feet

Answer 2

The velocity is the derivative of the position function s(t).

First find the derivative of s(t):

s'(t) = v(t) = -40/6t + 17/6

Next, the acceleration is the derivative of velocity function v(t). This is also the second derivative of s(t).

s''(t) = v'(t) = a(t) = -40/6

Determining the position of the shot when the velocity is 0: first, determine the time when velocity is 0.

0 = -40/6t + 17/6

0 = -40t + 17 (multiply both sides by 6)

40t = 17 (add 40t to both sides)

t = 17/40 (divide both sides by 40)

Now find position s(t) at t= 17/40:

s(17/40) = -20/6 * (17/40)² + 17/6 * 17/40 + 3

s(17/40) = 1729/480

Therefore, the position when velocity is 0 is 1729/480 or 3.602 feet.