"An object is rolled up an incline. If the object is 2.75 m up the incline after 4.50s and rolling back down at a velocity of 1.90 m/s, what is the acceleration of the object?"
Is this a case of final velocity - initial velocity?? It seems like an easy problem, but I don't know how to solve it. Thanks!
acceleration and velocity are two different things.
try using..
d = vt + (1/2)a(t^2)
remember: the object is rolling down for this equation
To solve this problem, we need to use the kinematic equation for the displacement of an object moving with constant acceleration:
\( \Delta x = v_i t + \frac{1}{2} a t^2 \)
where:
\( \Delta x \) is the displacement,
\( v_i \) is the initial velocity,
\( a \) is the acceleration, and
\( t \) is the time.
In this case, the object rolls up the incline, so the displacement is given as 2.75 m. The initial velocity, \( v_i \), is not given directly, but we know that the object is rolling back down at a velocity of 1.90 m/s. Since the object is rolling back down, the initial velocity, \( v_i \), would be the negative of the final velocity, \( v_f \), which is -1.90 m/s.
Substituting these values into the equation, we can solve for the acceleration:
\( 2.75 = (-1.90) \cdot 4.50 + \frac{1}{2} a \cdot (4.50)^2 \)
Simplifying the equation:
\( 2.75 = -8.55 + 10.125 a \)
Rearranging the equation to solve for 'a':
\( a = \frac{2.75 + 8.55}{10.125} \)
\( a = \frac{11.3}{10.125} \)
\( a \approx 1.12 \, \text{m/s}^2 \)
So, the acceleration of the object is approximately 1.12 m/s².
Note: Make sure to always pay attention to the signs when dealing with directions. The negative sign in front of the initial velocity accounts for the fact that the object is rolling back down the incline.