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A Year 12 exam features the following problem:

The acceleration of a body moving in a straight line is given by $a = 4v^2.$ We're given that $v = e$ when $x=1$.

The problem is to find the velocity of the body when $x = 2$.

I can't solve it.

We can get from $a=4v^2$ to $v = \frac{1}{A-4t}$ pretty easily, where $A$ is an unknown constant. What now?

goblin GONE
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1 Answers1

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Using what you already have, $v(t) = \frac{1}{A-4t}$, we integrate this expression to arrive at $$x(t) = \int\frac{dt}{A-4t} = -\frac{1}{4}\log|A-4t| + B$$ We manipulate this to arrive at an expression for $t(x)$ to plug into $v(t)$ to obtain a relation $v(x)$ which we can then evaluate. After some manipulation, we arrive at

$$t(x)=\frac{A\pm e^{4(B-x)}}{4}$$

Plugging this into $v(t) =\frac{1}{A-4t}$ leads to $A$ cancelling out and we obtain:

$$v(x) =\pm\frac{1}{e^{4(B-x)}} = \pm e^{4(x-B)}$$

With our condition that $v(1)=e$, we have:

$$e = e^{4-4B} \Rightarrow 1 = 4-4B\Rightarrow B=\frac{3}{4}$$

where we have dropped the negative case because $e$ is positive and $e^x$ is always positive as well.

Thus, the complete expression is $$v(x) = e^{4x-3}$$

We plug in $x=2$ and obtain $v(2) = e^5$.

Tom
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  • I should add that the given answer is $e^5$. Maybe an arithmetical error crept in somewhere? – goblin GONE Nov 20 '16 at 14:17
  • Yup! I forgot to cancel a $4$ when plugging into $v(t)$. I've edited the answer and it indeed agrees with your given answer now. Thanks for pointing that out! – Tom Nov 20 '16 at 14:22