2
$\begingroup$

Graph of a train (the body) is provided which starts from rest. What is the velocity after the train has 10 metres displacement?

enter image description here

Thats the only things provided for the question, please help me out here, I find that more variables will be required but the book states that no more variables are required.

The answer key says that the answer is 10 m/s only! I am totally unable to understand how.

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ What have you tried? What equations of motion do you know? (do you have a textbook which discusses this?) . $\endgroup$
    – Carl Witthoft
    Commented May 22, 2014 at 13:44
  • $\begingroup$ I have tried all possible combination of the equations I knew, @CarlWitthoft . The main problem that arises is that there are just too many variable quantities that are unknown, the eqns I tried were the main 3 only, and I think the question is probably to be done using only graphs, as it was in that specific section(By saying that, I mean that possibly, not many eqns of motion are needed to solve this), cheers and please help me if you can :) $\endgroup$
    – Nib
    Commented May 22, 2014 at 13:51
  • $\begingroup$ @user47047 The 3 basic equations you used can only be used when the accelration is constant, which here is not the case. $\endgroup$
    – Shubham
    Commented May 22, 2014 at 14:01
  • $\begingroup$ ya, youre right @Shubham $\endgroup$
    – Nib
    Commented May 22, 2014 at 14:07
  • $\begingroup$ But its not really a homework, I'm just doing the questions from the book for enrichment, @jinawee $\endgroup$
    – Nib
    Commented May 22, 2014 at 14:31

3 Answers 3

Reset to default
1
$\begingroup$

We know that $v^2-u^2=2ax$ is valid for constant acceleration. So, what are we waiting for? Lets find how this apply this.

We will use $v^2-u^2=2ax$ for a $very$ small time(or distance) for which we will say acceleration is constant.

$$v_0^2-u_0^2=2a_0x_0$$

$$v_1^2-u_1^2=2a_1x_1$$

$$v_2^2-u_2^2=2a_2x_2$$ $$\cdots$$

$$v_f^2-u_f^2=2a_fx_f$$

Note that $v_0=u_1$, $v_1=u_2$ as the intervals are directly after the time where previous equation was applied.

Add all of them.

$$v_f^2-u_0^2=2(a_0x_0+...a_fx_f)$$

Now look at the graph. Each term in right hand side represents area of the very small part of graph :enter image description here

Add all rectangles for complete area. Hence, right hand side is twice of area and as initial speed is $0ms^{-1},$ we get, $v_f^2=2\times\text{Area}$

Improve this answer
$\endgroup$
5
  • $\begingroup$ You can generalize this result.$2\times$ Area under $a-x$ graph is $v^2-u^2$ $\endgroup$
    – evil999man
    Commented May 22, 2014 at 14:36
  • $\begingroup$ whats v-subscript-f ? $\endgroup$
    – Nib
    Commented May 22, 2014 at 14:41
  • $\begingroup$ @user47047 Final speed. f stands for final $\endgroup$
    – evil999man
    Commented May 22, 2014 at 14:41
  • $\begingroup$ So, final speed comes out to be 180 m/ sec ? $\endgroup$
    – Nib
    Commented May 22, 2014 at 14:42
  • $\begingroup$ @user47047 It is $v^2$ and hence, answer is $\sqrt{180} m/s$ $\endgroup$
    – evil999man
    Commented May 22, 2014 at 14:44
1
$\begingroup$

The work done (per unit mass) after it has traveled a distance $x$ is the area under the acceleration curve between $0$ and $x$. $$W = 6 x - \frac{x^2}{10}$$

This work goes into kinetic energy (per unit mass) which is $K=\frac{1}{2} v^2$. Equating the two will give the velocity as a function of position

$$ v(x) = \sqrt{12 x - \frac{x^2}{5}} $$

Appendix

Area under acceleration curve $a(x) = 6 \left(1-\frac{x}{30}\right)$

Sketch

Area of rectangle$x \, a(x) = 6x-\frac{x^2}{5}$

Area of triangle $\frac{1}{2} x (6-a(x)) = \frac{x^2}{10}$

Total Area $ W = 6x-\frac{x^2}{5} + \frac{x^2}{10} = 6x-\frac{x^2}{10} $

Improve this answer
$\endgroup$
6
  • $\begingroup$ How did you reach W = 6x - x^2/10 ? $\endgroup$
    – Nib
    Commented May 22, 2014 at 14:45
  • $\begingroup$ @user47047 Formula for trapezium area, I guess. $\endgroup$
    – evil999man
    Commented May 22, 2014 at 14:47
  • $\begingroup$ But we dont have a trapezium here, @Awesome ! $\endgroup$
    – Nib
    Commented May 22, 2014 at 14:49
  • $\begingroup$ @user47047 He is talking about a general $x$ is between. Just plug in $x=30m$ $\endgroup$
    – evil999man
    Commented May 22, 2014 at 14:49
  • $\begingroup$ @user47047 Its the same area concept as Awesome used, Work is given by mass $times$ the area of the accelration displacement graph(You could, if you want, see it from $W=Fs$, and $F=ma$). Also, any work done on a body goes to change its kinetic energy. $\endgroup$
    – Shubham
    Commented May 22, 2014 at 14:59
0
$\begingroup$

According to the graph, accelration is linearly dependent on displacement. Now, assuming the motion is rectillinear(along a straight line, and quite justified for 30 m displacement of a train).

It is not very difficult to find the equation of this straight line (intercept form).

And then, you'll need some calculus to find velocity $\nu$ in terms of displacement $x$, which involves the following substitution : $$a=\nu\frac{d\nu}{dx}$$

Now transfer $dx$ to the other side and integrate. Try and see if it helps.

Improve this answer
$\endgroup$
17
  • 1
    $\begingroup$ Are you sure we need to use integration, I am only in 9th standard and we havent been taught integration or differentiation yet, and I dont really know how to do it! cheers, Nalin and pls see if any other method exists. :) $\endgroup$
    – Nib
    Commented May 22, 2014 at 13:52
  • 1
    $\begingroup$ And I also havent been taught Calculus yet :) $\endgroup$
    – Nib
    Commented May 22, 2014 at 13:53
  • 1
    $\begingroup$ $a=\color\red{v}\dfrac{dv}{dx}$ $\endgroup$
    – evil999man
    Commented May 22, 2014 at 14:12
  • 1
    $\begingroup$ @Shubham An argument can be made using averages and area. $\endgroup$
    – evil999man
    Commented May 22, 2014 at 14:17
  • 1
    $\begingroup$ @Awesome , sorry for not replying, but, the book uses a v-t graph and we just find s by the area of trapezium and substitute time to find the 3rd eqn $\endgroup$
    – Nib
    Commented May 22, 2014 at 14:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.