The answer to number 8 is; a = 0.6 m/s², the car's acceleration is in the positive (uphill) direction
The answer to number 9; is; -0.275 m/s². The negative sign indicates that Rohith's acceleration acceleration is in the opposite direction of his initial velocity.
8. To find the car's average acceleration, use the equation below for acceleration:
a = vf - vi / t
Where:
a is the acceleration
vf is the final velocity (4.5 m/s uphill)
vi is the initial velocity (-3.0 m/s downhill, but since uphill is chosen as the positive direction, it's -(-3.0) m/s, which becomes +3.0 m/s)
t is the time (2.5 s)
Input the values and calculate:
a = 4.5 m/s - 3.0 m/s / 2.5 s
a = 1.5 m/s / 2.5 s
a = 0.6 m/s²
Therefore, the car's average acceleration is 0.6 m/s² in the positive (uphill) direction.
9. To find Rohith's average acceleration during the 10.0 seconds, use the formula below for acceleration:
a = vf - vi / t
Where:
a is the acceleration.
vf is the final velocity (0.75 m/s).
vi is the initial velocity (3.5 m/s).
t is the time (10.0 s).
Input the values:
a = 0.75 m/s - 3.5 m/s / 10.0 s
a = -2.75 m/s / 10.0 s
a = -0.275 m/s²
Rohith's average acceleration during the 10.0 seconds is -0.275 m/s². The negative sign indicates that his acceleration is in the opposite direction of his initial velocity.