1) A person walks first at a constant speed of 5.10 m/s along a straight line fr

Question

1) A person walks first at a constant speed of 5.10 m/s along a straight line from point Ato point B and then back along the line from B toA at a constant speed of 2.90m/s.

(a) What is her average speed over the entire trip?

(b) What is her average velocity over the entire trip?

The answer that I got for both (a) and (b) were 4 m/s by adding thetwo values and then dividing by 2 but the homework site I'm onkeeps on telling me that it's the wrong answer and that I am"within 10% of the correct value" which means I'm close but I can'tfigure out how to do it any other way.

2) A 50.0 g Super Ball traveling at27.0 m/s bounces off a brick wall andrebounds at 20.0 m/s. A high-speed camerarecords this event. If the ball is in contact with the wall for3.00 ms, what is the magnitude of theaverage acceleration of the ball during this time interval? (Note:1 ms = 10-3 s.)

This one just plain stumps me. I tried the equation of averageacceleration ( v / t ) and then converting the milliseconds intoseconds but the answer is still wrong.

Explanation / Answer

(a) average speed = total distance / total time So lets say that the person walks 10 meters each way. total distance = 2*10m = 20m total time = distance/speed1 +distance/speed2 = (10m / 5.1m/s) + (10m /2.9m/s) = 5.4s dist/time = 20/5.4 = 3.7m/s (b) avg velocity= displacement / time displacement = 0 so avg velocity = 0 m/s Question 2: EDIT: sorry forgot to answer it lol hmm so a= v/t = vf - vi / t Make sure to assign directions for the velocity. vi = +27 and vf= -20 [because they're inopposite directions] so v= -20 - 27 = -47m/s t= .003s a = -47 / .003 = -1.6*104m/s2

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