An object moving with uniform acceleration has a velocity of 15.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 1.60 s later is -5.00 cm, what is its acceleration?
V = d / t = (-5-3)1.6 = -5m/s.
a = (Vf - Vo) / t,
a = (-5 - 15) / 1.6 = -12.5m/s^2.
Correction: V = -5cm/s, a = 12.5cm/s^2.
To find the acceleration of the object, we can use the kinematic equation:
x = x₀ + v₀t + (1/2)at²
Where:
x is the final position
x₀ is the initial position
v₀ is the initial velocity
t is the time
a is the acceleration
In this problem, we are given:
x₀ = 3.00 cm
v₀ = 15.0 cm/s
t = 1.60 s
x = -5.00 cm
Plugging in these values into the kinematic equation, we get:
-5.00 cm = 3.00 cm + (15.0 cm/s)(1.60 s) + (1/2)a(1.60 s)²
Now, let's solve for a:
-5.00 cm - 3.00 cm - (15.0 cm/s)(1.60 s) = (1/2)a(1.60 s)²
-8.00 cm - (24.0 cm) = (1/2)a(2.56 s²)
-32.0 cm = 1.28 a s²
Now, let's solve for a:
a = (-32.0 cm) / (1.28 s²)
a = -25 cm/s²
Therefore, the acceleration of the object is -25 cm/s².