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Mathematics 23 Online
OpenStudy (anonymous):

Please help me out. :( A car moving at a constant velocity of 14 m/s passes a gasoline station. Two seconds later, another car leaves the gasoline station and accelerates at the constant rate of 2 m/s^2. How soon will the second car overtake the first?

OpenStudy (anonymous):

my answer is 11.583 s and 2.417 s. is it correct?

ganeshie8 (ganeshie8):

show your work also so that others can avoid duplicating your effort :)

OpenStudy (anonymous):

okay. :)

OpenStudy (anonymous):

denotation: 1st Car = Honda 2nd Car = Toyota Given: dT = 2 m/s^2 vH = 14 m/s tT = tH - 2 dT = dH 1/2 at^2 = vt tT = 11.583 seconds tT = 2.417 seconds tT = tH - 2 tH = 13.583 seconds tH = 4.417 seconds ANSWER: tT = 11.583 seconds and 2.417 seconds

ganeshie8 (ganeshie8):

very nice try :)

ganeshie8 (ganeshie8):

2 m/s^2 this is the acceleration of second car right ? (Toyota)

OpenStudy (anonymous):

yes.. that's it..

ganeshie8 (ganeshie8):

So position equations for the cars can be given like below : First car(Hyundai) : \[\large S_h = 14t\] Second car(Toyota) : \[\large S_T = \dfrac{1}{2}2(t-2)^2 = (t-2)^2\]

ganeshie8 (ganeshie8):

When the second car overtakes the first car, the position of both cars would be same

ganeshie8 (ganeshie8):

So simply set both equations equal to eachother and solve \(\large t\)

ganeshie8 (ganeshie8):

\[\large S_h = S_T \] \[\large 14t = (t-2)^2 \]

ganeshie8 (ganeshie8):

Solve \(\large t\)

OpenStudy (anonymous):

i got two answers on that...

ganeshie8 (ganeshie8):

only one answer makes sense, throw away the other one

ganeshie8 (ganeshie8):

Considering the fact that the second car has started "2" seconds after the first car, which answer is invalid ?

OpenStudy (anonymous):

0.225 s.

ganeshie8 (ganeshie8):

yes thats the one you need to throw away, whats the correct answer and what does that represent exactly ?

OpenStudy (anonymous):

17.775 s. :)

ganeshie8 (ganeshie8):

thats right ! what does 17.775s represent ?

ganeshie8 (ganeshie8):

is it the time after First car or the time after Second car ?

OpenStudy (anonymous):

time after 2nd car..

ganeshie8 (ganeshie8):

nope

ganeshie8 (ganeshie8):

we have defined "t" as the time after the first car has started, right ?

ganeshie8 (ganeshie8):

So 17.775s corresponds to the time after the first car

OpenStudy (anonymous):

ahh.. okay.. im wrong. huhu

ganeshie8 (ganeshie8):

basically the Second car overtakes the First car exactly after 17.775s the First car starts So the first car travels for 17.7755s and the second car travels for 15.7755s before overtaking

OpenStudy (anonymous):

ahh.. so, what's the final answer?

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