Question

An arrow leaves a bow at 42 m/s. Its velocity is 34 m/s when it hits its target. Calculate distance it traveled in 2.4 seconds.

Answer 1

91.29 m

Explanation:

$\underline{\bigstar\:\textsf{Given:}}$

• Initial velocity of arrow when leaves a bow (u) = 42 m/s
• Final velocity of arrow when hits its target (v) = 34 m/s
• Time (t) = 2.4 s

$\underline{\bigstar\:\textsf{To\:find:}}$

• Distance the arrow traveled (s) = ?

$\underline{\bigstar\:\textsf{Solution:}}$

Let's take analysis of the situation first !

An arrow leaves a bow at 42 m/s (u), it's velocity after hitting it's target is 34 m/s (v) and it took 2.4 seconds now we have to calculate the distance it traveled.

Let's find now !

We know that, $\sf{\blue{acceleration(a)\:=\:\dfrac{v\:-\:u}{t}}}$

$\longrightarrow$ a = $\sf{\dfrac{34\:-\:42}{2.4}}$

$\longrightarrow$ a = $\sf{\dfrac{-8}{2.4}}$

$\longrightarrow$ a = -3.33 m/s²

So, the acceleration of arrow is -3.33 m/s.

Now, we also remember Newton's 3rd equation of motion i.e, v² = u² + 2as.

So, $\sf{\blue{distance(s)\:=\:\dfrac{v^2\:-\:u^2}{2a}}}$

$\implies\:s\:=\:\sf{\dfrac{(34)^2\:-(42)^2}{2\:×\:-3.33}}$

$\implies\:s\:=\:\sf{\dfrac{1156\:-\:1764}{-6.66}}$

$\implies\:s\:=\:\sf{\dfrac{-608}{-6.66}}$

$\implies\:s\:=\:\sf{\dfrac{608\:×\:100}{666}}$

$\implies$ s = 91.29 meters

${\large{\boxed{\rm{\therefore \: Distance\:traveled\:by\:the\:arrow\:is\:91.29\:m}}}}$

Hope it helped you dear...

Answer 2

$\red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Distance traveled by arrow = 91.29m

$\red{\mid{\underline{\overline{\textbf{explanation}}}\mid}}$

Acceleration is the rate of change of velocity of a body. It is represented by the expression :

$\bigstar \: \frac{v - u}{t}$

In this expression :

• v = Final velocity
• u = Initial velocity
• t = time taken

Newtons third equation of motion says that square of final velocity of a body is equals to square of its initial velocity and double of product of acceleration and displacement. It is represented by the expression :

$\bigstar \: {v}^{2} = {u}^{2} + 2as$

In this expression :

• v = Final velocity
• u = Initial velocity
• a = acceleration
• s = displacement

$\blue{\mid{\underline{\overline{\textbf{</p><p>Coming to question :-}}}\mid}}$

Important keywords :

• Final velocity of arrow= v = 34m/s
• Initial velocity of arrow = 42m/s
• Time taken to hit the target = 2.4s
• Acceleration = ?
• Distance = ?

Now acceleration of arrow will be :

$\mapsto \: \frac{34 - 42}{2.4}$

$\mapsto \: \frac{ - 8}{2.4} = - 3.33$

Therefore acceleration of arrow is -3.33m/s

Now after inputting the values in third equation of motion we get :

$s = \frac{ {v}^{2} - {u}^{2} }{2a}$

$\mapsto \: s = \frac{ {34}^{2} - {42}^{2} }{2 \times 3.33}$

$\mapsto \: s = \frac{608}{6.66} = 91.29$

Therefore distance traveled by arrow is 91.29m

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BY SCIVIBHANSHU

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