Kinematics Problem: constant acceleration, motion in a line

[MariaComeHere]

Homework Statement

An object moives along the straight line EFGH with a constant acceleration. It takes the object seven seconds to travel between E and F, three seconds to travel the 66m between F and G, and four seconds to travel between G and H, where it comes to rest. What is the deceleration of the object? What is the velocity of the object at E?

Relevant Equations

The SUVAT equations

I've been attempting to solve this problem for three days now. I have thrown away my old attempts (like, scrumpled up into the bin), but my old attempts involved:

Trying to set up simultaeneous equations relating the journeys between EH and FG to find the deceleration, but the reason why this didn't work is because I did not have enough common information between EHand FG. There does not seem to be a way to get enough information. If I examine the journey between E and H, then I know that t = 3 + 4 + 7 = 14 and that v = 0 (since it comes to rest at H) and am trying to find a, but this doesn't give me enough information to work anything out. Moreover, between F and G I have that t = 3, s = 66 and I'm trying to find a. Once again, I don't have enough information.

I could try setting up simultaeneous equations, but the issue with this is that I don't have two bits of common information. The only common bit of information between the journeys EH and FG is t. I do not see how I am supposed to solve for the deceleration and velocity of the object at E.

Could someone walk me through the solution so that I can learn from it?

 

[berkeman]

Welcome to the PF.

Interesting problem. It does seem underconstrained. I drew a triangle with vertical axis v(t) and horizontal axis t. The v(t) for the object is the hypotenuse of the triangle starting at v(E) on the vertical axis, and linearly decreasing until zero at v(H). The points F and G are easy to plot on that graph.

But v(E) does not seem constrained by anything. It can be higher to start, but that just makes the deceleration greater. Are you sure there is nothing else mentioned in the problem statement?

 

[MariaComeHere]

Welcome to the PF.

Interesting problem. It does seem underconstrained. I drew a triangle with vertical axis v(t) and horizontal axis t. The v(t) for the object is the hypotenuse of the triangle starting at v(E) on the vertical axis, and linearly decreasing until zero at v(H). The points F and G are easy to plot on that graph.

But v(E) does not seem constrained by anything. It can be higher to start, but that just makes the deceleration greater. Are you sure there is nothing else mentioned in the problem statement?

I have been drawing a straight line while solving the problem. I agree that I don't see how I can get enough information. There is nothing else mentioned - I copied it exactly.

Also, the answer that the book gives is 4 ms^-2 for the deceleration, and 56 ms^-1 for the velocity at E.

 

[berkeman]

Also, the answer that the book gives is 4 ms^-2 for the deceleration, and 56 ms^-1 for the velocity at E.

Well, 56/14 = 4, so those answers are consistent. But I still don't see anything constraining v(E). I tried doing a Google search for the problem, to see if I could see any other constraints, but no joy so far.

Where is this problem from? Is it in a textbook, or a problem set from your instructor. Is there any way you can ping a TA or the instructor to ask if there is something missing from the problem statement?

 

[PAllen]

Hmm, it seems to me the problem is pretty straightforward, at least as I interpret it. You have uniform deceleration over a straight line. There are two unknowns of the motion, v(E) and the deceleration. There are two constraints, expressed as follows in terms of distance from E as function of time ( s(t) ), and v as function of time. The general form of these (v and s as functions of t) should be obvious, and fully determined given t measured from 0 when the object is at E, and s being 0 at t=0.

Then the two constraints are:

s(10) - s(7) = 66

v(14) = 0

This fully determines the two unknowns, and agrees with the stated solution.

 

[berkeman]

s(10) - s(7) = 66

Ack, I missed the 66m in the original question. Thanks!

 

[PeroK]

Homework Statement: An object moives along the straight line EFGH with a constant acceleration. It takes the object seven seconds to travel between E and F, three seconds to travel the 66m between F and G, and four seconds to travel between G and H, where it comes to rest. What is the deceleration of the object? What is the velocity of the object at E?
Homework Equations: The SUVAT equations

Could someone walk me through the solution so that I can learn from it?

Another approach is to consider average velocity. For constant acceleration, average velocity is always half way between the initial and final velocities for any period of the motion.

In this case, we have $v_E, v_F, v_G$ and $v_H = 0$. And, a constant deceleration of $-a$, say, where I'll take $a$ to be positive. With SI units assumed:

$v_H = v_G - 4a$, hence $v_G = 4a$

$v_G = v_F - 3a$, hence $v_F = 7a$

$v_F = v_E - 7a$, hence $v_E = 14a$

Now, the average velocity between $F$ and $G$ is $5.5a$, which is half way between $v_G = 4a$ and $v_F = 7a$.

And, taking $66m/3s = 22m/s$ gives the average velocity between $F$ amd $G$.

Hence $5.5a = 22$ and $a = 4m/s^2$.