Question

Physics quiz 11 th-12th grade helpp

Answer 1

5. The jogger's velocity is a constant 3.55 m/s between t = 4 s and t = 8 s.

6. Given a linear plot of velocity, the acceleration is determined by the slope of the line. Take any two points on the part of the plot after t = 8 s - for instance, we see it passes through (8 s, 3.5 m/s) and (10 s, 4 m/s) - and compute the slope:

(4 m/s - 3.5 m/s)/(10 s - 8 s) = (0.5 m/s)/(2 s) = 0.25 m/s^2

7. This amounts to finding the area between the velocity function and the time axis and between t = 4 s and t = 8 s. During this time, the velocity is 3.5 m/s. The time interval lasts 4 s. So the distance covered is

(3.5 m/s)*(4 s) = 14 m

8. After 4 seconds, Jimmy's speed decreases from 30.0 m/s to 27.2 m/s, so his acceleration (assuming it was constant) was

a = (27.2 m/s - 30.0 m/s)/(4 s) = -0.200 m/s^2

It's unclear what is meant by "rate of acceleration", since the acceleration is itself a rate. But maybe they just mean to ask for the acceleration, or possibly the magnitude?

Answer 2

#5). At t=6 seconds, the jogger's instantaneous speed is 3.5 m/s ... just as it has been since t=4 seconds.

#6). At t=10 seconds, her acceleration is 0.25 m/s² .

#7). From t=4 seconds until t=8 seconds, her speed is constant at 3.5 m/s. So, in that period of 4 seconds, she covers (3.5 m/s x 4 sec) = 14 meters.