Two Dimensional Motion (Seemingly Simple)

[learnitall]

Homework Statement

An object starts from rest at time t=0.00 s and moves in the +x direction with constant acceleration. The object travels 13.0 m from time t=1.00s to time t=2.00 s. What is the acceleration of the object?
a) 5.20 m/s^2
b) 10.4 m/s^2
c) 8.67 m/s^2
d) 6.93 m/s^2
e) 12.1 m/s^2

The correct answer is c)

Homework Equations

1. v = v₀ + at
2. x= x₀ + v₀t + 1/2 at
3 v²= v₀² + 2a(Δx)
4 Δx= 1/2 (v + v₀)t
5. v_av-x = (x₂ - x₁)/(t₂-t₁)

The Attempt at a Solution

The problem says that the object starts from rest and assuming my system starts at zero i can use equation 2 to get:

13 m = 0 + 0 + 1/2 a (2)^2
which gives me:
a = 6.5 m/s^2

my next attempt was using the fact that the object traveled more than 13 m because we must take into account the distance traveled in the first second ( from t=0.00 s to t=1.00 s)

To solve for this distance which I called 'd' I tried using equation 2 3 and 4 by substituting 'x' for '13 + d'. Eventually I reach a point where i have to use quadratic formula to solve for 'd' and substitute it back into one of my chosen equations. It gets really complicated and I can't seem to find a nice answer for 'a'. Can anyone tell me what's a good way to solve for 'a'? My way seems too difficult to solve in 5 min (this was a question on an old exam). How do you get 8.67 m/s^2 for 'a'?!

 

[gneill]

Homework Statement

An object starts from rest at time t=0.00 s and moves in the +x direction with constant acceleration. The object travels 13.0 m from time t=1.00s to time t=2.00 s. What is the acceleration of the object?
a) 5.20 m/s^2
b) 10.4 m/s^2
c) 8.67 m/s^2
d) 6.93 m/s^2
e) 12.1 m/s^2

The correct answer is c)

Homework Equations

1. v = v₀ + at
2. x= x₀ + v₀t + 1/2 at
3 v²= v₀² + 2a(Δx)
4 Δx= 1/2 (v + v₀)t
5. v_av-x = (x₂ - x₁)/(t₂-t₁)

The Attempt at a Solution

The problem says that the object starts from rest and assuming my system starts at zero i can use equation 2 to get:

13 m = 0 + 0 + 1/2 a (2)^2
which gives me:
a = 6.5 m/s^2

Well you know that can't be right because the object is going to have some initial velocity when it begins the 13m portion of its trip, and the time taken to cover just that portion is not 2 seconds (what is it?).

my next attempt was using the fact that the object traveled more than 13 m because we must take into account the distance traveled in the first second ( from t=0.00 s to t=1.00 s)

More importantly you need to take into account the speed it gains during the first part of its trip! You really don't care how FAR it traveled in that first second since how far it traveled before doesn't matter to the only portion of the trajectory that is of interest: the final 13 meters.

To solve for this distance which I called 'd' I tried using equation 2 3 and 4 by substituting 'x' for '13 + d'. Eventually I reach a point where i have to use quadratic formula to solve for 'd' and substitute it back into one of my chosen equations. It gets really complicated and I can't seem to find a nice answer for 'a'. Can anyone tell me what's a good way to solve for 'a'? My way seems too difficult to solve in 5 min (this was a question on an old exam). How do you get 8.67 m/s^2 for 'a'?!

As I say, the initial distance doesn't matter; The speed v_o when it begins the final 13m matters. Write an expression for the just the final part of the journey supposing that you know v_o, Δs, and Δt. What's an expression for v_o?

 

[learnitall]

As I say, the initial distance doesn't matter; The speed v_o when it begins the final 13m matters. Write an expression for the just the final part of the journey supposing that you know v_o, Δs, and Δt. What's an expression for v_o?

Sorry for the late response. I had classes all day and got caught up with other hw.

So I thought about what you said but instead of considering velocities I set up a simpler equation (after all this was an exam so there HAD to be easier way).

I considered displacement:
x(2)-x(1) = 13 m

1. 1/2 (a) (2 s)^2 - 1/2 (a) (1 s)^2 = 13 m

2. 2a - 1/2 a = 13

3. 3/2 a = 13

4. 3a = 26 m/s^2

5. a= 8.66 m/s^2

 

[gneill]

Yes, that works too. Quite elegant, in fact.