Question

the distance travelled by a boy in the nth second is given by the expression (2+3n).find the initial velocity and the acceleration and also find final velocity in 2 second?

Answer 1

The distance travelled by a boy in the nth second is given by the expression $(2 + 3n)$. find the initial velocity and the acceleration and also find final velocity in 2 second?

By definition:

$$v = \frac {\Delta l}{\Delta t}$$

$\Delta l$ – distance travelled in time $\Delta t$

In our case, distance travelled by a boy in the nth second is given by the expression $(2 + 3n)$, so

$$v (n) = \frac {2 + 3 n}{1} = 2 + 3 n$$

$v(n)$ – average velocity during nth second.

Acceleration equals:

$$a = \frac {\Delta v}{\Delta t} = \frac {v (n + 1) - v (n)}{n + 1 - n} = 2 + 3 (n + 1) - (2 + 3 n) = 3$$

Answer: Acceleration equals $3\frac{\text{units}}{\text{second}^2}$.

Average velocity during first second equals:

$$v (1) = 2 + 3 = 5$$

And he moved with acceleration $a = 3$.

$$v _ {0} + 3 * 1 = v _ {1}$$

$v_{0}$ – the initial velocity

$v_{1}$ – the final velocity in 1st second

He moved with constant acceleration, so:

$$v (1) = \frac {v _ {0} + v _ {1}}{2} = 5$$

Therefore, we have system of equations:

$$\left\{ \begin{array}{l} v _ {0} + 3 = v _ {1} \\ \frac {v _ {0} + v _ {1}}{2} = 5 \end{array} \right.$$

Substitute 1 to 2:

$$(v _ {0} + v _ {0} + 3) = 2 * 5$$ $$2 v _ {0} = 10 - 3 \quad v _ {0} = 7 / 2$$

Answer: the initial velocity equals $7 / 2 \frac{\text{units}}{\text{second}}$.

Average velocity during 2 second equals:

$$v (2) = 2 + 3 * 2 = 8$$

And he moved with acceleration $a = 3$.

$$v _ {0} + 3 * 1 = v _ {1}$$

$v_{0}$ – the initial velocity in $2^{\mathrm{nd}}$ second

$v_{1}$ – the final velocity in $2^{\mathrm{nd}}$ second

He moved with constant acceleration, so:

$$v (2) = \frac {v _ {0} + v _ {1}}{2} = 8$$

Therefore, we have system of equations:

$$\left\{ \begin{array}{l} v _ {0} + 3 = v _ {1} \\ \frac {v _ {0} + v _ {1}}{2} = 8 \end{array} \right.$$

From second: $v_{0} = 16 - v_{1}$

Substitute 1:

$$16 - v _ {1} + 3 = v _ {1}$$ $$2 v _ {1} = 19 \quad v _ {1} = 19 / 2$$

Answer: final velocity in 2 second equals $\frac{19}{2}$ units

units