Where and when does the car catch the truck?

  • Thread starter adoule
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  • #1
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Homework Statement



A truck starts from rest and accelerates at 1 M/S*2 . 4 s later, a car accelerates from rest at the same starting point with an acceleration of 2.7 M/S*2 .

Where and when does the car catch the truck?

Homework Equations





The Attempt at a Solution


i have tried many equations but nothing is working :s could someone please help
 
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Answers and Replies

  • #2
lewando
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A truck starts from rest and accelerates at . s later, a car accelerates from rest at the same starting point with an acceleration of .
Please clarify what you mean by "acceleration of ."

i have tried many equations but nothing is working :s could someone please help!
Please show what you have done so that help can be provided.
 
  • #3
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Please clarify what you mean by "acceleration of ."


Please show what you have done so that help can be provided.
i clarified!
 
  • #4
lewando
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Thanks, adoule. So what have you tried so far?
 
  • #5
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Thanks, adoule. So what have you tried so far?
i have tried s=ut + (1/2)at^2

In 4 seconds, truck travels a distance of 0 (4) + (1/2) (1) (4)^2 = 0 + 8 = 8

now, when car starts, truck is already 8 m ahead of it.

Suppose they meet at time lapse of t seconds after car starts.
So, car travels 8 m more than the truck in t seconds.
distance travelled by car in t seconds = dist travelld by truck in t seconds + 8

0 (t) + (1/2) (2.7) (t)^2 = 0 ( t) + (1/2) (1) (t)^2 + 8

==> (t)^2 = (160) / 17
t = 3.06

hence, car catches truck at 4 + 3.06 = 7.06 seconds after truck starts.
distance travelled by car = 0 (7.06) + (1/2) (4) ( 160/17) = 18.8 metres
 
  • #6
lewando
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i have tried s=ut + (1/2)at^2

In 4 seconds, truck travels a distance of 0 (4) + (1/2) (1) (4)^2 = 0 + 8 = 8

now, when car starts, truck is already 8 m ahead of it.
I'm with you at this point.
Suppose they meet at time lapse of t seconds after car starts.
So, car travels 8 m more than the truck in t seconds.
distance travelled by car in t seconds = dist travelld by truck in t seconds + 8
still with you...
0 (t) + (1/2) (2.7) (t)^2 = 0 ( t) + (1/2) (1) (t)^2 + 8
The RHS is not correct. You need to take into account the velocity of the truck (not zero!)
 
  • #7
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I'm with you at this point.


still with you...


The RHS is not correct. You need to take into account the velocity of the truck (not zero!)
ok so i corrected the velocity of the truck it gives me this
0 (t) + (1/2) (2.7) (t)^2 = 4 ( t) + (1/2) (1) (t)^2 + 8
and then it gives me an equation like this
0= 0.85t^2- 4t-8
and the anwser is 6.2192121 but its still nor correct :s
and sorru about the other thread :p
 
  • #8
lewando
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The answer you are getting, 6.2, is right for the equation, but you need to account for the 4s time shift. Need to add 4 s to it (you were trying to solve for a specific time interval τ, which began at t = 4).
 

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