Question

as a result of the collision, the first object goes from a velocity of $1.00 \frac{m}{s}$ to $-0.50 \frac{m}{s}$, and the second object goes from a velocity of $-0.50 \frac{m}{s}$ to $1.00 \frac{m}{s}$. assume that the collision takes $0.01$ s.

find the acceleration of each object using the equation $a = \frac{\delta v}{t}$, where $\delta v$ is the change in the velocity of an object, and $t$ is the time it takes the velocity to change.

calculate the acceleration of each object.
enter your answers in the boxes.

$a_1 = \square \frac{m}{s^2}$

$a_2 = \square \frac{m}{s^2}$

Explanation:

Step1: Calculate acceleration of first object

The initial velocity \( v_{i1} = 1.00 \, \frac{m}{s} \), final velocity \( v_{f1} = -0.50 \, \frac{m}{s} \), time \( t = 0.01 \, s \). Change in velocity \( \Delta v_1 = v_{f1} - v_{i1} = -0.50 - 1.00 = -1.50 \, \frac{m}{s} \). Acceleration \( a_1 = \frac{\Delta v_1}{t} = \frac{-1.50}{0.01} \).
\[ a_1 = -150 \, \frac{m}{s^2} \]

Step2: Calculate acceleration of second object

Initial velocity \( v_{i2} = -0.50 \, \frac{m}{s} \), final velocity \( v_{f2} = 1.00 \, \frac{m}{s} \), time \( t = 0.01 \, s \). Change in velocity \( \Delta v_2 = v_{f2} - v_{i2} = 1.00 - (-0.50) = 1.50 \, \frac{m}{s} \). Acceleration \( a_2 = \frac{\Delta v_2}{t} = \frac{1.50}{0.01} \).
\[ a_2 = 150 \, \frac{m}{s^2} \]

Answer:

\( a_1 = -150 \, \frac{m}{s^2} \)
\( a_2 = 150 \, \frac{m}{s^2} \)