Acceleration and velocity questioneasyneed helpby vicsic Tags: acceleration, distance, easy, speed, velocity 

#1
Sep1711, 09:25 PM

P: 13

1. The problem statement, all variables and given/known data
A car is waiting at a stoplight. Just as the light turns green and the car starts to accelerate, a truck passes at a constant velocity in the next lane. The truck passes at a velocity 25.0 m/s while the car accelerates 5.00 m/s^2 both in the same direction, Calculate the distance travelled by the car before it overtakes the truck? Car's speed as it overtakes the truck? 2. Relevant equations I am a little confused on this one. 3. The attempt at a solution I think you have to use the formula vf^2=Vi^2 +2ad but am not sure. Do you combine 2 formulas? any help would be appreciated 



#2
Sep1711, 09:57 PM

P: 112





#3
Sep1711, 10:12 PM

P: 13





#4
Sep1711, 10:19 PM

P: 183

Acceleration and velocity questioneasyneed help 



#5
Sep1711, 10:24 PM

P: 112

Recheck your equation. This is how it goes:
X = Xi + Vi(t) + (.5)Ax(t^2) Xc = 0 + 0(t) + (.5)(5.00)(t^2) = 2.5t^2 XT = 0 + 25.0(t) + (.5)(0)(t^2) = 25.0t SO....2.5t^2 = 25.0t.......THEN.....t=10. 



#6
Sep1711, 10:32 PM

P: 13

vi=0 m/s a=5.00 m/s^2 Truck v=25.0 m/s assuming the distances are the same at the beginning distance would equal zero as they are stopped using the formula d=Vi(t)+1/2at^2 0=0t+1/2(5.00)t^2 0=1/2(5.00)t^2 0=5.00t^2 0=t^2 t=0 plug that back into the formula D=Vf(t)1/2at^2 D=25(t)1/2(5.00)t^2 D=0 this is my work for this and it is not getting me the right answer which in the back of the book says 250m for A and 50 m/s for B any help please. 



#7
Sep1711, 10:42 PM

P: 112

The reason we set the distance equal to one another (NOT TO 0, because we dont know the distance) is that we are finding WHEN the car meets up with the trucks position. By solving for t, as I did above, you do that. 



#8
Sep1711, 10:44 PM

P: 112

Also, the reason you were getting t=0 is that you were solving for when the cars were at distance 0...which is t = 0.




#9
Sep1711, 10:52 PM

P: 13

d=vi(t) + 1/2at^2 how did u get all of those x's into it and 0's for the x's thanks for all the help! 



#10
Sep1711, 11:06 PM

P: 112

d(the position of x from Xi aka the distance) = Xi(Initial position of x which is 0 and why they probably left it out) + Vi(t)(which is initial velocity, 0 for the car cause it is at rest in the beginning, and 25 is given for the truck) + 1/2Axt^2(I put the x because it just means Acceleration in the x direction, but you probably haven't went into acceleration in the y direction yet so that is why it is tripping you up. Just think that Ax is equal to what you have for a, Also, A is 0 for the truck because the velocity is constant, and 5.0 is given for the car.) That should cover everything, hope that helps. 



#11
Sep1811, 08:57 AM

P: 13

:) 


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