Question

A body starts from rest and moves with a uniform acceleration of 2 m s/ 2 until it travels a distance of 625 m. Find its velocity.

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Answer 1

Given :

Initial Velocity = 0 m/s

Acceleration = 2 m/s²

Distance covered = 625 m

To Find :

Velocity (v)

Solution :

Solving via first calculating the time using second equation of motion and then using third equation of motion to find final velocity, just a little different approach ^^"

We have,

s = distance = 625 m

u = 0 m/s

t (to be calculated)

a = 2 m/s²

So this gives the perfect set for using second equation of motion.

Third Equation of Motion :

\bold{\large{\boxed{\mathtt{s=\:ut\:+\:{\dfrac{1}{2}\:\:at^2}}}}}

s=ut+

2

1

at

2

Block in the available data,

➠ \mathtt{625=0\:\times\:t\:+\:{\dfrac{1}{2}\:2\:t^2}}625=0×t+

2

1

2t

2

➠ \mathtt{625=0\:+\:{\dfrac{1}{2}\:2t^2}}625=0+

2

1

2t

2

➠ \mathtt{625\:=\:{\dfrac{1}{2}\:\:2t^2}}625=

2

1

2t

2

➠ \mathtt{625=\:{\dfrac{2t^2}{2}}}625=

2

2t

2

➠ \mathtt{1250\:=\:2t^2}1250=2t

2

➠ \mathtt{\dfrac{1250}{2}\:=\:t^2}

2

1250

=t

2

➠ \mathtt{625\:=\:t^2}625=t

2

➠ \mathtt{\sqrt{625}\:=\:t}

625

=t

➠ \mathtt{25\:=t}25=t

\bold{\sf{\therefore{\underline{time\:taken\:=\:25\:second}}}}∴

timetaken=25second

Now, simply use the third equation of motion and calculate v.

\bold{\large{\boxed{\mathtt{v\:=\:u\:+\:at}}}}

v=u+at

Block in the available data,

➠ \mathtt{v=0\:+\:2\:\times\:25}v=0+2×25

➠ \mathtt{v=\:50}v=50

\bold{\sf{\therefore{\underline{Final\:Velocity\:(v)\:=\:50\:ms^-1}}}}∴

FinalVelocity(v)=50ms

−

1

Answer 2

Answer:

$\huge\fbox\red{Plz refer the attachment}$