An object moving with uniform acceleration has a velocity of 10.0 cm/s in the positive x-direction when its x-coordinate is 2.91 cm. If its x-coordinate 1.85 s later is −5.00 cm, what is its acceleration?
But who does acceleration in cm^2/s
Changing to meters
at t = 0
ui = 0.10 m/s
xi = 0.0291 m
at t = 1.85 s
u = 0.10 + 1.85 a
x = 0.0291 + 0.10 (1.85) + .5 a (1.85)^2 = -0.05 so solve this for a
To find the acceleration of the object, we can use the following equation of motion:
x = x0 + v0*t + (1/2)*a*t^2
where:
x is the final displacement
x0 is the initial displacement
v0 is the initial velocity
t is the time
a is the acceleration
In this case, we are given the initial velocity (v0 = 10.0 cm/s), the initial displacement (x0 = 2.91 cm), the final displacement (x = -5.00 cm), and the time (t = 1.85 s).
We can solve the equation for acceleration (a) by rearranging the equation:
a = (2*(x - x0 - v0*t)) / t^2
Now, let's substitute the given values into the equation:
a = (2*(-5.00 cm - 2.91 cm - 10.0 cm/s * 1.85 s)) / (1.85 s)^2
Calculating the expression:
a = (2*(-5.00 cm - 2.91 cm - 18.5 cm)) / (1.85 s)^2
a = (-14.82 cm) / (3.4225 s^2)
a ≈ -4.327 cm/s^2
Therefore, the acceleration of the object is approximately -4.327 cm/s^2.