Question

Actual velocity data for a lion pursuing prey are shown in the figure. Estimate the initial acceleration of the lion, at 2 s, 4 s, and the distance traveled by the lion between 0 s and 8 s? Express your answer to one significant figure and include the appropriate units. I'm not sure if my answers are right. I got 0 m/s for initial but my response is "Enter your answer using dimensions of acceleration." I thought it was m/s and I tried m/s^2 but that is wrong too.. Can someone please help?

Answer 1

The acceleration is the slope of this graph, at the points asked draw tangents

Answer 2

The acceleration (slope) at 2 sec is about 2.5 m/\(s^2\)

Answer 3

????

Answer 4

what part did you not understand?

Answer 5

How to get tangent line?

Answer 6

you draw it

Answer 7

I have one attempt left for my answer for an acceleration of the lion at 4s. Can someone help?1?!

Answer 8

Just to explain what @aaronq was getting at: remember, acceleration is the change in velocity over time. This means if you plot velocity against time on a graph like you have here, the SLOPE of the graph represents the acceleration. If we want to find the instantaneous acceleration, we have to find the change in velocity between two points on the graph that are veeeeeeery close together (otherwise it's just an average acceleration). The tangent line that @aaronq is talking about represents the slope of the graph at a SPECIFIC point, making that slope the instantaneous acceleration.

Now, on to your question...

Even though your starting velocity is 0 m/s according to the graph, the acceleration around this point is not. If you look at the graph from t=0 to about t=1, the graph looks pretty straight, so the acceleration is roughly constant during this time and the average and instantaneous accelerations are the same. Over 1 s, the velocity changes from 0 to 5 m/s, so the acceleration is 5 m/s^2 (since a = change in v / change in t).

Now try a similar strategy for 2 s and 4 s!

Answer 9

I gave up and lost points on the first answer of o which Mastering Physics gave me was 6 m/s^2 for the first answer and then for 2 s it is 2 m/s^2 but for 4 seconds i can't figure it out.. Please somebody help?!

Answer 10

it's not 0 and it's not at a constant rate between 2 s and 4 s so what is the ratio?

Answer 11

it looks like it is slowing down in velocity

Answer 12

You're right! That means the acceleration is smaller during this time. For the other time points, you still do the same thing, and assume the line between the times you choose is straight. So for 4 s, I'm looking at the average velocity from 3 to 5 s and assuming that is a good representation of the instantaneous velocity at 4 s.

The velocity at 3 s looks to be about 12 m/s, and the velocity at 5 s is about 14 m/s. Using the equation for acceleration, this means the acceleration is roughly 1 m/s^2. Does that make sense?

Answer 13

Yeah, thank you for explaining further for me :)