Please help! I don't know how to solve these problems!
1
A runner dashes from the starting line (x = 0) to a point 130 m away and then turns around and runs to a point 11 m away from the starting point in 21 seconds. To the nearest tenth of a m/s what is the average speed?
2
What is the runner's average velocity in the previous problem?
3
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
Sure, I can help you solve these problems step by step:
1. To find the average speed, we divide the total distance traveled by the total time taken. In this case, the runner traveled 130 m + 11 m = 141 m in 21 seconds. So, the average speed can be calculated as follows:
Average speed = Total distance / Total time
= 141 m / 21 s
≈ 6.7 m/s (rounded to the nearest tenth)
Therefore, the average speed of the runner is approximately 6.7 m/s.
2. Average velocity is different from average speed because velocity takes into account direction. In this case, the runner runs a total distance of 130 m + 11 m = 141 m, but ends up 11 m away from the starting point.
Since the total displacement is negative (-11 m), the runner's average velocity can be calculated as follows:
Average velocity = Displacement / Total time
= -11 m / 21 s
= -0.5 m/s
Therefore, the runner's average velocity is -0.5 m/s.
3. To find the average acceleration, we need to calculate the change in velocity and divide it by the time taken. In the first 10 seconds, the car accelerates from 0 m/s to 32.6 m/s, and in the next 5 seconds, it slows down to 14.8 m/s.
The change in velocity can be calculated as:
Change in velocity = Final velocity - Initial velocity
= 14.8 m/s - 32.6 m/s
= -17.8 m/s
The average acceleration can be calculated as:
Average acceleration = Change in velocity / Total time
= -17.8 m/s / 15 s
≈ -1.2 m/s^2 (rounded to the nearest tenth)
Therefore, the car has an average acceleration of approximately -1.2 m/s^2.