Over a time interval of 2.37 years, the velocity of a planet orbiting a distant star reverses direction, changing from +23.4 km/s to -15.7 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.
To find the total change in velocity of the planet during the time interval, we need to subtract the initial velocity from the final velocity.
Given:
Initial velocity, u = +23.4 km/s
Final velocity, v = -15.7 km/s
We first need to convert the units from km/s to m/s for the calculation.
1 km/s = 1000 m/s
So, the initial velocity in m/s is:
u = +23.4 km/s * 1000 m/s/km = +23400 m/s
And the final velocity in m/s is:
v = -15.7 km/s * 1000 m/s/km = -15700 m/s
(a) To find the total change in velocity, we subtract the initial velocity from the final velocity:
Δv = v - u
Δv = -15700 m/s - (+23400 m/s)
Δv = -39100 m/s
The negative sign indicates that the velocity has reversed direction.
Therefore, the total change in velocity of the planet is -39100 m/s.
(b) To find the average acceleration, we use the formula:
Average acceleration = (change in velocity) / (time interval)
Given:
Time interval, t = 2.37 years
We need to convert the time interval from years to seconds for the calculation.
1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
So, the time interval in seconds is:
t = 2.37 years * 365 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute
t ≈ 74,797,200 seconds
Now we can calculate the average acceleration:
Average acceleration = (Δv) / (t)
Average acceleration = (-39100 m/s) / (74,797,200 s)
Calculating this gives us the average acceleration.
Therefore, the average acceleration of the planet during this interval is approximately -0.000522 m/s². The negative sign indicates that the acceleration is directed opposite to the initial velocity.