Find magnitude of car's acceleration

[Rijad Hadzic]

Homework Statement

While strolling downtown on a Saturday afternoon you stumble across an old car show. As you are walking along an alley toward a main street, you glimpse a particularly stylish Alpha Romero pass by. Tall buildings on either side of the alley obscure your view, so you see the car only as it passes between the buildings. Thinking back to your physics class, you realize that you can calculate the cars acceleration. You estimate the width of the alley way between the two buildings to be 4 m. The car was in view for .5s. You also heard the engine rev when the car started from a red light, so you know the alpha romero started from rest 2s before you first saw it. Find the magnitude of its acceleration

Homework Equations

[itex] V_{ox} + a_xt = V_x [/itex]
[itex] \Delta x = V_xt - (1/2) a_xt^2 [/itex]

The Attempt at a Solution

So since the car starts from 0 velocity, its final velocity =

[itex] V_x = a_x(t_2) [/itex]
where [itex] t_2 = 2 s[/itex]

its initial velocity [itex] V_{ox} = 2\Delta x + a_xt_1^2 / (2t_1) [/itex] where [itex] t_1 = .5s [/itex]

is = to the final velocity first equation, so the two are =

[itex] 2\Delta x + a_xt_1^2 / (2t_1) = a_x(t_2) [/itex]

now solving

[itex] a_x (2t_1)(t_2) = 2\Delta x + a_x t_1^2 [/itex]
[itex] a_x(2t_1)(t_2) - a_xt_1^2 = 2\Delta x [/itex]

[itex] a_x [ (2t_1)(t_2) - (t_1)^2 ] = 2\Delta x [/itex]
[itex] a_x = (2\Delta x ) / (2t_1)(t_2)-(t_1^2) [/itex]

[itex] = a_x = 8 m / (1s)(2s) - (.5s)^2 = 8m / (1.75 s)^2 = 4.6 m/s^2 [/itex]but my book is telling me my answer is aboout 4 seconds.

My answer is closer to 5 seconds.

Does anyone know what I did wrong? Maybe the way I set up my eq's?

 

[BvU]

Check ##\Delta x = V_xt - (1/2) a_xt^2## : do you really think an accelerating car would go backwards after ##2v\over a## seconds ?

 

[Rijad Hadzic]

Check ##\Delta x = V_xt - (1/2) a_xt^2## : do you really think an accelerating car would go backwards after ##2v\over a## seconds ?

I think I understand now. To answer your question, no lol.

Using [itex] \Delta x = V_{ox}t + (1/2)a_xt^2 [/itex] would make sense because the velocity and acceleration would be consistent, right?

and my [itex] V_{ox} [/itex] would be = to my first equations [itex] V_x [/itex] anyways so I can set them equal, right? Sin [itex] V_{ox} [/itex] is just another way of writing [itex] V_x [/itex]

 

[Rijad Hadzic]

Anyways did the problem again and got 3.6, and they say the answer is ABOUT 4 which is pretty vague but I think I did it correctly now using

[itex] \Delta x = V_{ox}t + (1/2)a_xt^2 [/itex]

where Vox = Vx in the first equation

 

[haruspex]

Anyways did the problem again and got 3.6, and they say the answer is ABOUT 4 which is pretty vague but I think I did it correctly now using

[itex] \Delta x = V_{ox}t + (1/2)a_xt^2 [/itex]

where Vox = Vx in the first equation

I get your answer if I ignore the length of the car. Never heard of Alfa Romero, but an Alfa Romeo Giulia is over 4m long. If it was in (at least partial) view for 0.5s then it traveled about 8m in that time, making the acceleration more like 7m/s².

 

[BvU]

I agree. Haru points out a weakness in the problem statement. Not clear if a single point on the car is in view for 0.5 s (which I think was intended -- so car travels 4m in 0.5 s) or any part of the car (as Haru explains).