I don't know if im doing these problems right, please help...
A car accelerates at a constant rate from zero to 32.6 m/s in 10 seconds and then slows to 14.8 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s2 during the 15 seconds
--would you do 32.6 + 14.8, and then divide by 15?
What was the acceleration during the first 10 seconds in the previous problem
--would you do 32.6 divided be 10?
Let V1 = starting velocity
Let V2 = ending velocity
The problem asks for the average acceleration for the 15 second period. The average acceleration would be the change in velocity over the time period.
dv/dt = (V2 - V1)/(15 seconds)
In this case, the starting velocity is zero.
To find the average acceleration during the 15 seconds, you need to calculate the total change in velocity and divide it by the total time.
Step 1: Calculate the change in velocity:
The car starts from a velocity of zero and accelerates to 32.6 m/s in 10 seconds, which gives a change in velocity of 32.6 m/s - 0 m/s = 32.6 m/s.
Then, the car slows down to 14.8 m/s in the next 5 seconds, which gives another change in velocity of 14.8 m/s - 32.6 m/s = -17.8 m/s.
To calculate the total change in velocity, add these two changes together: 32.6 m/s + (-17.8 m/s) = 14.8 m/s.
Step 2: Calculate the average acceleration:
Divide the total change in velocity by the total time taken:
Average acceleration = (14.8 m/s) / (15 s) ≈ 0.99 m/s² (rounded to the nearest tenth).
For the second question, you were correct. To find the acceleration during the first 10 seconds, you divide the change in velocity during that time by the time interval. In this case, the change in velocity was 32.6 m/s and the time interval was 10 seconds.
Acceleration during first 10 seconds = 32.6 m/s / 10 s = 3.26 m/s².