Question

Tiger Shroff brings his horse to rest in 1. 5s. If the braking force acting, produces a deceleration of 5m/s², calculate the initial velocity of the horse in Km/hr.

Answer 1

Answer:

27 km/h

Explanation:

As per the provided information in the given question, we have :

• Final velocity (v) = 0 m/s [Comes to rest]
• Time taken (t) = 1.5 s
• Acceleration (a) = – 5m/s²

We've been asked to calculate the initial velocity of the horse in km/hr.

⠀⠀⠀⠀⠀⠀⠀⠀» According to the question, we have to values of v , t , a ; in order to calculate the u, we'll have to apply a proper equation of motion through which we can find the value of u. There are mainly three equations of motion,

• $\boxed{ \sf{v = u + at} }$
• $\boxed{\sf{ s = ut + \dfrac{1}{2}at^2}}$
• $\boxed{ \sf{v^2 - u^2 = 2as}}$

Here,.

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes distance

Basically, if we have a look at the equations then we can easily say that the most suitable equation which can be applied is the first equation of motion. So by using the first equation of motion,

$\dashrightarrow \quad \rm { v = u + at}$

$\dashrightarrow \quad \rm { 0 = u + (-5)(1.5)}$

$\dashrightarrow \quad \rm { 0 = u + (-7.5)}$

$\dashrightarrow \quad \rm { 0 = u - 7.5}$

$\dashrightarrow \quad \rm { 0 + 7.5 = u }$

$\dashrightarrow \quad \underline{\boxed{ \bf {7.5 \; m \: s^{-1} = u }}}$

∴ The initial velocity of the horse in m/s is 7.5 m/s.

― Now, we have to convert it into km/h, in order to that so, we have to multiply the initial velocity in m/s with ¹⁸/₅.

$\dashrightarrow \quad \rm { u = 7.5 \; m \: s^{-1} }$

$\dashrightarrow \quad \rm { u = \Bigg \{ 7.5 \times \dfrac{18}{5} \Bigg \} \; km \: h^{-1} }$

$\dashrightarrow \quad \rm { u = \Bigg \{ 1.5 \times 18 \Bigg \} \; km \: h^{-1} }$

$\dashrightarrow \quad \underline{\boxed{ \bf {u = 27 \; km \: h^{-1} }}}$

∴ The initial velocity of the horse in km/h is 27 km/h.

$\rule{200}2$

Answer 2

$\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}$

Tiger Shroff brings his horse to rest in 1.5 s . If the braking force acting, produces a deceleration of 5m/s², calculate the initial velocity of the horse in Km/hr.

$\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}$

Initial velocity of the horse = 27 km/hr

$\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}$

$\green{ \large \underline{ \mathbb{\underline{GIVEN : }}}}$

• Tiger Shroff brings his horse to rest in 1.5 s .

• The braking force acting, produces a deceleration of 5m/s² .

$\green{ \large \underline{ \mathbb{\underline{TO \: FIND : }}}}$

• Initial velocity of the horse in Km/hr .

$\green{ \large \underline{ \mathbb{\underline{EQUATION \: OF \: MOTION\: USED: }}}}$

$\purple{ \longrightarrow \boxed{ \tt v =u + at }}$

$\green{ \large \underline{ \mathbb{\underline{SOLUTION: }}}}$

Final velocity ( v ) = 0 $\sf m {s}^{ - 1}$

Acceleration ( a ) = - 5 $\sf m {s}^{ - 2}$

Deceleration means negative acceleration or acceleration in opposite direction .

Time taken ( t ) = 1.5 s

$\displaystyle \bf \implies 0 = u + ( - 5)1.5 \\ \\ \displaystyle \bf \implies0 = u - 7.5 \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \bf \implies u - 7.5 = 0 \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \bf \implies u = 7.5 \: m {s}^{ - 1} \: \: \: \: \:$

Now , to convert m/s into km/h we will multiply it with 18/5 .

$\displaystyle \bf \implies u = 7.5 \times \frac{18}{5} \: km \: {h}^{ - 1} \\ \\ \displaystyle \bf \implies u = 1.5 \times 18\: km \: {h}^{ - 1} \\ \\ \displaystyle \bf \implies u = 27\: km \: {h}^{ - 1} \: \: \: \: \: \: \: \: \: \: \: \:$

$\green{ \large \underline{ \mathbb{\underline{KNOW\:MORE: }}}}$

$\purple{\boxed{\begin{array}{l} \textsf{Equation of motion : } \\ \\ \textsf{v = u + at} \\ \\ \displaystyle\textsf{s = ut + \sf\frac{1}{2}a {t}^{2} } \\ \\ \sf {v}^{2} - {u}^{2} = 2as \end{array}}}$