Question

A bicycle initially m
5.0 m s-accelerates for
will be its final velocity ?
initially moving with a velocity
celerates for 5 s at a rate of 2 ms. What will be it's final velocity
​

Answer 1

Correct question :

A bicycle initally moving with a velocity of 5 m/s. It accelerated for 5 s at a rate of 2 m/s^2. what will be it's final velocity at the end of 5 second.

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Final\:velocity=15\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: velocity(u) = 5 \: ms \\ \\ \tt: \implies Acceleration(a) = 2 \: {ms}^{2} \\ \\ \tt: \implies Time(t) = 5 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Final \: velocity(v) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 5 + 2 \times 5 \\ \\ \tt: \implies v = 5 + 10 \\ \\ \green{\tt: \implies v = 15 \: m/s} \\ \\ \bold{Alternate \: method : } \\ \tt: \implies s = ut + \frac{1}{2}{at}^{2} \\ \\ \tt: \implies s = 5 \times 5 + \frac{1}{2} \times 2 \times 25 \\ \\ \tt: \implies s =25 + 25 \\ \\ \tt: \implies s = 50 \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {v}^{2} = {5}^{2} + 2 \times 2 \times 50\\ \\ \tt: \implies {v}^{2} = 25 + 200 \\ \\ \tt: \implies {v}^{2} = 225 \\ \\ \tt: \implies v = \sqrt{225} \\ \\ \green{\tt: \implies v = 15 \: m/s}$

Answer 2

AnswEr:-

Final velocity (v) = 15 m/s

$\rule{200}1$

Correct question :-

A bicycle initially moving with a velocity of 5 m/s accelerates for 5 s at a rate of 2 m/s².What will be it's final velocity?

Given :-

• Initial velocity (u) = 5 m/s
• Acceleration(a) = 2 m/s²
• Time(t) = 5 s

To find :-

• Final velocity (v) = ?

Solution :-

We know formula:-

$\star\: {\boxed{\rm{\blue{s = ut + \dfrac{1}{2} at^2 }}}}$

$\dashrightarrow\sf s = 5(5) + \dfrac{1}{2}\times 2(5)^2 \\\\\dashrightarrow\sf s = 25 + \dfrac{1}{2}\times 2 \times 25 \\\\\dashrightarrow\sf s = 25 + \dfrac{1}{2}\times 50 \\\\\dashrightarrow\sf s = 25 + 25 \\\\\dashrightarrow{\underline{\boxed{\textsf{\textbf{s = 50 m }}}}}\: \star$

$\rule{100}{2}$

We also know,

$\star\:{\boxed{\rm{\green{v^2 = u^2 + 2as }}}}$

$\dashrightarrow\sf v^2 = (5)^2 + 2 \times 2 \times 50 \\\\\dashrightarrow\sf v^2 = 25 + 200 \\\\\dashrightarrow\sf v^2 = 225 \\\\\dashrightarrow\sf v = \sqrt{225} \\\\\dashrightarrow{\underline{\boxed{\bold{v = 15 \: m/s^2 }}}}\: \star$

∴ Final velocity = 15 m/s²