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?
in 1.85 s, it has moved -7.91 cm
... find the average velocity ... distance / time
the final velocity is ... (2 * average) - initial
acceleration is ... (final velocity - initial velocity) / time
To find the acceleration of the object, we can use the kinematic equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d) of an object:
vf^2 = vi^2 + 2ad
In this case, the initial velocity (vi) is 10.0 cm/s in the positive x-direction, the final velocity (vf) is unknown, the displacement (d) is the difference between the final and initial x-coordinates, and the time interval (t) is 1.85 s.
First, we can find the displacement (d):
d = xf - xi
= (-5.00 cm) - (2.91 cm)
= -7.91 cm
Now, we need to calculate the final velocity (vf) using the initial velocity (vi), displacement (d), and time interval (t):
vf^2 = vi^2 + 2ad
vf^2 = (10.0 cm/s)^2 + 2(a)(-7.91 cm)
vf^2 = 100.0 cm^2/s^2 - 15.82(a)
Next, we substitute the known values into the equation:
(10.0 cm/s)^2 + 2(a)(-7.91 cm) = vf^2
100.0 cm^2/s^2 + 2(-7.91 cm)(a) = vf^2
Now, we need to find vf by finding the square root of both sides of the equation:
vf = sqrt(100.0 cm^2/s^2 + 2(-7.91 cm)(a))
Finally, we need to solve for the acceleration (a). By rearranging the equation, we can isolate a on one side:
vf = sqrt(100.0 cm^2/s^2 + 2(-7.91 cm)(a))
vf^2 = 100.0 cm^2/s^2 + 2(-7.91 cm)(a)
2(-7.91 cm)(a) = vf^2 - 100.0 cm^2/s^2
a = (vf^2 - 100.0 cm^2/s^2) / (2(-7.91 cm))
Now, substitute the values of vf and solve for a:
a = [(vf^2) - 100.0 cm^2/s^2] / [-15.82 cm]
a = [(sqrt(100.0 cm^2/s^2 + 2(-7.91 cm)(a))^2) - 100.0 cm^2/s^2] / [-15.82 cm]
By substituting the given values into the equation, we can calculate the acceleration.