Physics, asked by satrox3187, 1 year ago

A marble rolling on a smooth floor has an initial velocity of 0.4m/s .if the floor offer a retardation of 0.02 m/s,calculate the time it will take to come at rest

Answers

Answered by ArchitectSethRollins
25

\underline {hello friend}
\color {green} solution :
Initial velocity (u) = 0.4 m/s

Acceleration (a) = - 0.02 m/s²

Final velocity (v) = 0 m/s. [because, the marble will come to rest .]

Time (t) = ?

\boxed{\boxed{using \: formula : v  = u + at}} \\  \\ \implies 0 = 0.4 + ( - 0.02 \times t) \\  \\ \implies  - 0.4 =  - 0.02t \\  \\ \implies t =  \frac{ - 0.4}{ - 0.02}  \\  \\ \implies t = 20 \: seconds

Therefore,

\boxed{\boxed{time = 20 seconds}}
\color{blue}hope \:  it  \: helps

Answered by aryanthestar005
11
u=0.4
a= -0.02
v=0 (as it comes to rest)
t=?
We will use the v=u+at
0=0.4+0.02(t)
-0.4=0.02(t)
-0.4
------- = t
0.02
t = 20 seconds
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