2

Can someone please help me in finding the formula used to get the answer to this question?

An object moves along $x$-axis. In any coordinate $x$, the acceleration is $a=x^4$ (SI units). If the object goes from rest in $x=1$m, what velocity it will get at $x=2$m? \begin{align*} A)\quad 4.59 \textrm{ m/s}\\ B)\quad 3.52 \textrm{ m/s}\\ C)\quad 2.47 \textrm{ m/s}\\ D)\quad 5.66 \textrm{ m/s}\\ E)\quad 1.41 \textrm{ m/s}\\ \end{align*}

Answer B is the correct answer

Stahl
  • 23,212
pedroh
  • 35

2 Answers2

5

If $v=\frac{dx}{dt}$ is the velocity at time $t$, we have $$\frac{dv}{dt}=x^4.$$ There is a standard trick for solving this kind of DE. Multiply both sides by $v$. We get $$v\frac{dv}{dt}=vx^4=x^4\frac{dx}{dt}.$$ Now we can integrate with repect to $t$. On the left, we get $\frac{1}{2}v^2$, and on the right we get $\frac{1}{5}x^5 +C$. Thus $$\frac{1}{2}v^2=\frac{1}{5}x^5 +C.$$ Use your initial condition to find that $C=-\frac{1}{5}$. Now we know $v^2$ at position $x$.

André Nicolas
  • 507,029
  • Thank you, finally I get to the answer! So a formula to this type of problem can be: 1/2*v^2=integrate(a,x)+C. First we replace v and x with the initial condition to get C. Then we replace C, and x with the second state of initial condition right? – pedroh Jun 16 '13 at 21:50
  • Yes, that's right. In your case we end up with $v^2=\frac{2}{5}(x^5-1)$, which is $12.4$ at $x=2$. Now take the square root. – André Nicolas Jun 16 '13 at 21:55
  • Yes I get it, thank you again! – pedroh Jun 16 '13 at 21:56
1

Hint: Think about the relationship between velocity and acceleration in terms of calculus.

Let $v(x)$ be the velocity at position $x$; $a(x)$ be the acceleration at position $x$:

$$\int a(x)\ dx=v(s)+C,\ a(x)=v(x)\frac{d}{dx}$$

  • I can't get to this answer. If the acceleration was according to time, v=integrate(a,t) will do it, but I can't understand with positions.. – pedroh Jun 16 '13 at 21:28