Question

a body moves over a horizontal surface with a velocity of 2 minute per second due to friction its velocity degrees uniformly at rate of 0.25 second squire how much time will it take to stop​

Answer 1

Correct question:-

A body moves over a horizontal surface with an initial velocity of 2 m/s. Due to friction, its velocity decreases uniformly at the rate of 0.25 m/s^{2}. How much time will it take to stop?

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

$\green{\tt{\therefore{Time\:taken=8\:sec}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: speed = 2 \: m/s \\ \\ \tt: \implies Acceleration = - 0.25 \: m/ {s}^{2} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Time \: taken \: to \: stop \: body =?$

• According to given question :

$\tt \circ \: Final \: speed = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 0 = 2 + ( - 0.25) \times t \\ \\ \tt: \implies - 2 = - 0.25t \\ \\ \tt: \implies t = \frac{ - 2}{ - 0.25} \\ \\ \green{\tt: \implies t = 8 \: sec}\\\\\ \blue{ \boxed{ \bold{Some \: related \: formula}}} \\ \orange{ \tt \circ\: \: s = ut + \frac{1}{2} a {t}^{2} } \\ \\ \orange{ \tt \circ \:\: {v}^{2} = {u}^{2} + 2as} \\ \\ \orange{ \tt \circ \:\: Speed = \frac{Distance}{Time} }$

Answer 2

${ \huge{ \bold{ \underline{ \underline{ \green{Question:-}}}}}}$

A body moves over a horizontal surface with a velocity of 2 minute per second. Due to friction its velocity degrees uniformly at rate of -0.25m/s² .. How much time will it take to stop ?

_______________

${ \huge{ \bold{ \underline{ \underline{ \orange{Answer:-}}}}}}$

✒Given : -

• Initial Velocity(u) = 2m/s
• Acceleration(a) = -0.25m/s²
• Final Speed(v) = 0m/s

✒To Find : -

• Time(t) = ?

✒Formula Used : -

$\leadsto\sf{{ \large{ \bold{ \bold{ \bold{ \red{v=u+at}}}}}}}$

✒Now ,

${ \large{ \bold{ \underline{ \underline{ \purple{According\:to\:the\:Question:-}}}}}}$

$\dashrightarrow\sf{v=u+at}$

$\dashrightarrow\sf{0=2+(-0.25)\times{t}}$

$\dashrightarrow\sf{-2=-0.25t}$

$\dashrightarrow\sf{t=\cancel\dfrac{-2}{-0.25}}$

$\leadsto\sf{{ \large{ \boxed{ \bold{ \bold{ \green{t=8\:sec.}}}}}}}$

⤳Therefore , time taken by the body to stop = 8 seconds ..