An object has a constant acceleration of 40 ft/sec2, an initial velocity of -20 ft/sec, and an initial position of 10 ft
well, you know that
a = 40
v = ∫a dt = 40t + a
s = ∫v dt = 20t^2 + at + b
use the given conditions to find a and b.
s(t)= (20t sqrd -20t +10)
Use the formula
s(t)= (1/2)a tsqrd + V0t + S0
To find the position of the object at any given time, you can use the following equation:
position = initial position + initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial position is given as 10 ft, the initial velocity as -20 ft/sec, and the acceleration as 40 ft/sec^2.
Let's calculate the position of the object at a specific time, say t = 5 seconds.
First, plug in the values into the equation:
position = 10 ft + (-20 ft/sec) * 5 sec + (1/2) * 40 ft/sec^2 * (5 sec)^2
Simplifying each term:
position = 10 ft + (-100 ft) + (1/2) * 40 ft/sec^2 * 25 sec^2
position = 10 ft - 100 ft + 20 ft/sec^2 * 25 sec^2
position = -90 ft + 500 ft
position = 410 ft
Therefore, at t = 5 seconds, the object's position is 410 ft.