A body moving with uniform acceleration has a velocity of 11.1 cm/s when its x coordinate is 4.6 cm. If its x coordinate 1.75 s later is −4.29 cm, what is the x-component of its acceleration? Answer in units of cm/s2. How would I set this up?
x(t)=xo+vix*time+1/2 a t^2
let toriginal=first sentence conditions.
V(t)=vo+at but at the first sentence, t is zero, so vo=11.1
so now go to the last position..
x(t)=xo+vix*time+1/2 a t^2
-4.29=0+11.1(1.75)-1/2 a (1.75)^2
solve for a.
To solve this problem, we can use the kinematic equations of motion. Let's break down the information given to set up the problem:
1. The body has a velocity of 11.1 cm/s when its x-coordinate is 4.6 cm. This gives us the initial velocity (u) and initial position (x₀).
u = 11.1 cm/s
x₀ = 4.6 cm
2. The body's x-coordinate 1.75 seconds later is -4.29 cm. This gives us the final position (x) and the time (t).
x = -4.29 cm
t = 1.75 s
We need to find the x-component of the body's acceleration (aₓ). The equation that relates initial velocity, final velocity, acceleration, and displacement is:
x = x₀ + ut + (1/2)at²
Since we are only concerned with the x-component, we can rewrite the equation as:
x = x₀ + uₓt + (1/2)aₓt²
Here, uₓ represents the x-component of initial velocity.
To find aₓ, we need to rearrange the equation:
aₓ = (2x - 2x₀ - 2uₓt) / t²
Now we can substitute the given values into the equation:
aₓ = (2(-4.29 cm) - 2(4.6 cm) - 2(11.1 cm/s)(1.75 s)) / (1.75 s)²
Calculating this expression will give us the x-component of the body's acceleration in units of cm/s².