Introductory kinematics word problem

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shaools
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Homework Statement



A competitor is aiming to complete a 1500m wheel-chair race in less than 4.0 minutes. After moving at a constant speed for exactly 3.5 minutes, there were still 240m to go. What must his acceleration before the remaining distance if he were to finish the race on time?

Given:
d1= 1260m [forward]
d2= 240m [forward]
t1= 210s
t2= 30s

Required: v1, v2 possibly?, and a


Homework Equations



v1 = d/t
and d = v1*(t+a(t)^2)/2 ?

The Attempt at a Solution



v1=1260m/210s
v1=6m/s

d = v1*(t+a(t)^2)/2
a= (2d/t^2)-v1*t
a= (2*240m/30s^2) - 6m/s*30s
a= something horribly incorrect

the answer at the back of the book is 0.13m/s^2. I am guessing that I am not approaching the second part of the question correctly. i tried it again with an equation using v2, but I am still getting weird answers.

help please !

also, I am sorry i don't know how to use latex :/
 
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Welcome to PF!

Hi shaools! Welcome to PF! :smile:

(don't bother about LaTeX for something like this … the SUB and SUP tags are fine, and they save a bit of server space also! :wink:)
shaools said:
d = v1*(t+a(t)^2)/2
a= (2d/t^2)-v1*t
a= (2*240m/30s^2) - 6m/s*30s

hmm … you're messy with brackets, which leads you to make mistakes

that middle line is wrong. :smile:
 
wow.

i managed to screw up the initial equation twice in that post. there's no way to edit posts is there?

i meant to post this equation for displacement and acceleration:

http://id.mind.net/~zona/mstm/physics/mechanics/kinematics/EquationsForAcceleratedMotion/Origins/Displacement/Image78.gif"

soo...

d = v1*t + (a*t^2)/2
a = 2d/(t^2) - v1*t

and then when i plug my values in, i get funky answers.
i think i might just be using the wrong equation, or i might be plugging in incorrect values. I am really not sure.

thank you for checking tho!
 
Last edited by a moderator:
ah, bested by bedmas

lol let me attempt this once more...

a= [2(d - vt)]/ t^2 ?
 
a = [2(240m - 6m/s*30s)] / 30^2
a = 0.13m/s^2

yup :).

thank you mr. fish !