Question

Rahul brings his horse to rest in 1. 5s If the braking force acting, produces a deceleration of
5m/s calculate the initial velocity of the horse in km/hr​
​

Answer 1

Answer:

27km/hr

Explanation:

t = 1.5s a = 5m/s

u = at = 5×1.5 = 7.5 m/s

now convet into km/hr

7.5 ×60×60/1000km/hr

Answer 2

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

Rahul 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

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

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

• Rahul 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 .

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

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

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