# What final velocity does the motorcycle attain?

## a motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a final velocity v. then t is increased so that the displacement is 3x. (don't have enough room on top)

Aug 22, 2017

Final velocity of motorcycle is $\sqrt{3}$ times initial velocity.

#### Explanation:

Data:
Initial velocity$= {v}_{i} = 0$
Final velocity$= {v}_{f} = v$
Acceleration$= a$
Distance$= s = x$

We know that:
${v}_{f}^{2} - {v}_{i}^{2} = 2 a s$
$\implies {v}^{2} - {0}^{2} = 2 a x$
$\implies {v}^{2} = 2 a x$

Now, data for second case when distance becomes $3 x$.

Initial velocity$= {v}_{i} = 0$
Final velocity=v_f=??
Acceleration$= a$
Distance$= s ' = 3 x$

Again,
${v}_{f}^{2} - {v}_{i}^{2} = 2 a s '$
$\implies v {f}^{2} - {0}^{2} = 2 a \left(3 x\right)$
$\implies v {f}^{2} = 3 \left(2 a x\right) = 3 {v}^{2}$
$\implies v f = \sqrt{3} v$
$\implies$ final velocity of motorcycle is $\sqrt{3}$ times initial velocity.

Aug 22, 2017

The final velocity is $1.732 \cdot v$.

#### Explanation:

I will assume that the total displacement, after the additional time period, is 3x. (The other possible interpretation I see is that the 3x is in addition to the original 1x for a total displacement of 4x.)

The calculations in item numbers 1 & 2 in the list below are analysing what we know about the first time interval t.

1. Using the kinematic formula
$s = \frac{u + v}{2} \cdot t$
we find that the displacement $x = \frac{0 + v}{2} \cdot t = \frac{v \cdot t}{2}$
2. Using the kinematic formula
$v = u + a \cdot t$
we find that $v = 0 + a \cdot t$ and from that we can rearrange to determine the acceleration:
$a = \frac{v}{t}$.

Now we will consider the time period in which the total displacement is 3x. We do not know length of that time period. So we look to the set of kinematic formulas to find a formula that will give us the final velocity (${v}_{f}$) without knowing the time.
Using the kinematic formula
${v}^{2} = {u}^{2} + 2 \cdot a \cdot s$
and plugging in our data (note, s = 3x)
${v}_{f}^{2} = {0}^{2} + 2 \cdot \left(\frac{v}{t}\right) \cdot \frac{3 \cdot v \cdot t}{2}$,
and simplifying
${v}_{f}^{2} = \cancel{2} \cdot \left(\frac{v}{\cancel{t}}\right) \cdot 3 \cdot \left(\frac{v \cdot \cancel{t}}{\cancel{2}}\right)$
we finally have
${v}_{f} = \sqrt{3 \cdot {v}^{2}} = 1.732 \cdot v$

If the total displacement in fact was 4*x, start with the paragraph below item 2 in that list, replace the 3x with 4x, and redo the remaining work.

I hope this helps,
Steve