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Velocity and acceleration of a runner

  • #1

Homework Statement



Dan starts from rest and started running for a distance of 6.0 x 101 m in 9 s.
a) What was Dan’s final velocity at this time?
b) What was the acceleration of Dan?


Homework Equations



a= final velocity - initial velocity / change in time


The Attempt at a Solution



Initial velocity = 0
t= 9 s
distance = 60 m


a= final velocity - initial velocity / change in time

the problem Im having is that i don't have acceleration or final velocity, so i don't know how i'm sopposed to plug into the formula

I was just hoping for a push in the right direction! thanks
 

Answers and Replies

  • #2
radou
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Dan is accelerating uniformly. Which relation tells us about the distance traveled in this case of motion?
 
  • #3
Dick
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You should probably assume he is accelerating at a constant rate during this time. Do you know any equations relating distance to acceleration?
 
  • #4
distance = (initial velocity)(change in time) + 1/2a (change in time)(change in time)

that is an equation relating to distance nd acceleration
 
  • #5
Dick
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Ok, so zero initial velocity. You have the distance, you have the change in time. So you should be able to find 'a', right?
 
  • #6
yes, i think i actually figured it out!

his acceleration would be 1 because v/t is 15/15 which is 1 m/s squared
and then you plug that in and his final velocity would be 22.9 but with sigdigs it would be 23

does that seem right?
 
  • #7
Dick
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Whoa! Slow down. I now have:

60 m=(1/2)*a*(9 sec)^2 and I'm asking myself what is 'a'? I don't see any v/t and I don't get 1.
 
Last edited:
  • #8
Okay, i got excited and started using my givens from a different questions, what i meant was:

you have to isolate for a so you have

distance/aceleration = (1/2)(81)
= 40.5
then you cross multiply

distance = acceleration (40.5)
distance/ 40.5 = a
60/40.5 = a
a = 1.481 m/s squared
 
  • #9
Dick
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Well done. So final velocity is?
 
  • #10
v = at
= (1.481 m/s)(9)
= 13.32 m/s
= sig digs 13 m/s
 
  • #11
Dick
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Seems fine to me. But I would write (1.481 m/s^2)*(9 s)=13.32 m/s. Keep the units straight.
 
  • #12
Thankyou so much

Do you think i need to have a direction since it's velocity? or if i say +13 m/s will that be fine?

Would you help me with one more question? it is

Toronto downtown and Scarborough Town Centre are 20.0 km apart. A cyclist leaves Toronto and heads for Scarborough at 20.0 km/h. A second cyclist leaves Scarborough for Toronto at exactly the same time at a speed of 15.0 km/h.
a) Where will the two cyclist meet between Toronto and Scarborough?
b) How much time passes before they meet (in minutes)?

I know the distance and two speeds and i am looking for T I was thinking maybe to useing a formula with distance velocity and time, and then isolating for time, but i don't know which formula
 
  • #13
Dick
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13 m/sec is fine. We are pretty clear on the direction, right? Ok, for the second one just use d=v*t. Let t be the unknown meeting time. What distance does each cyclist travel in time t? When they meet, what will the sum of those distances be? Can you solve the resulting equation for t?
 
  • #14
Okay so you have t = d/t
t = 20 km/ 15 km/h
t = 1.33 hours

So they will meet after 1.33 hours
 
  • #15
Dick
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Nope. Slow down again. If the time is T, how far does the first cyclist travel?
 
  • #16
I don't know the question doesn't say.d = vt
= 20 km/h time

I don't know what to doooo
 
  • #17
I don't know the distance,they don't tell you, and there is no way to find out. It there?
 
  • #18
Dick
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As you said, the first bicyclist travels (20km/h)*T. The second one travels (15km/h)*T. Since T is the meeting time, if I add those two together, what do I get?
 
  • #19
displacement? so the read formula would be

t = distance 2 -distance 1 /velocity 2 -velocity 1

= 20km / 20km/h - 15 km/h
= 20km/5km/h
= 4 hours??
 
  • #20
Dick
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Forget formulas! Answer my question! I want a number of km! What is the sum? How much distance did the riders cover together?
 
  • #21
(20km/h)*T. The second one travels (15km/h)*T add them together 35 km/g
 
  • #22
that was sopposed to say 35 km/s
 
  • #23
thankyou for all you help but i have to go.

THANKYOU SO MUCH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 
  • #24
Dick
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that was sopposed to say 35 km/s
Noo. It was supposed to say 35km/h. Ok, so (35 km/h)*T=?. What is ??. How much distance did they travel altogether?
 
  • #25
Dick
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So do I. Think it over again.
 

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