Question

Find instantaneous velocites of Tennis player at:
a) .50s b)2.0s c) 3.0s d)4.5s
I will include a picture and how I did a) I just dont understand how to do the rest.

Answer 1

Here is how i did A)

Answer 2

@Data_LG2 Do you think you can help me with this problem? I am going to leave to school in about 20 minutes and I have been stuck on this all night. Physics is new to me

Answer 3

You know velocity is the slope of the graph right?

For the first one, since the velocity is constant from t=0 to t=1, the velocity at t=0.5 will be the same as their average velocity..

For the second one, since the velocity is constant from t=1 to t=2.5, the velocity at t=2s will be the same as their average velocity.

For the 3rd one, since the velocity is constant from t=2.5 to t=4, the velocity at t=3s will be the same as their average velocity.


For the 4th one, since the velocity is constant from t=4 to t=5, the velocity at t=4.5 s will be the same as their average velocity.

Answer 4

sorry late response I was away

Answer 5

its okay, I am just trying to figure out this physics. haha they say its easy but when I read about it I am just as confused
Thank you and let me see what you wrote.

Answer 6

Yes I did know the velocity is the slope, and from the graph shown.

Answer 7

just tell me if you're confused with anything (:

Answer 8

Okay let me understand or type this then.

a) .50s => 1m-0m / 4s-0s = .25 s \[.50s => \frac{ 1m-0m }{ 4.0s-0s } = .25\]


wait I did something wrong

Answer 9

t=0 to t=1 not t=0 to t=4

Answer 10

for letter a

Answer 11

for a) I did \[.50s=\frac{2m }{ .50s } = 4.0 m/s\]

for b) from what you said @Data_LG2 I did
\[b) 2.0s=\frac{ -2m-4m }{ 2.5s-1s } => -4.0 m/s\] Is this one right? because its final minus inital for both the x= distance and t=time.

Answer 12

@ Data thank you, I think I got it...I checked the back of the book and it only gives the answers and I got them. thank you.

Answer 13

OH MY GOODNESS! I reallly have to leave now. Have a wonderful day/night I have class

Answer 14

\(\sf \large a. v_{ins @ .50s} = \frac{(4-0)m}{(1-0)s}\)

\(\sf \large b. v_{ins @ 2s} = \frac{(-2-4)m}{(2.5-1)s}\)

\(\sf \large c. v_{ins @ 3s} = \frac{(-2+2)m}{(4-2.5)s}\)

\(\sf \large d. v_{ins @ 4.5s} = \frac{(0+2)m}{(5-4)s}\)