Physics, asked by sergiojr5166, 1 day ago

. A car initially at rest starts moving with a constant acceleration of
0.5 ms-2 and travels a distance of 36 m. Find its final velocity.

Answers

Answered by Anonymous
35

Given :

  • Car is initially at rest [u =0]

  • Acceleration of the body is 0.5 m/s² [a = 0.5 m/s²]

  • Distance travelled is 49m [s = 49m]

To find :

  • Final velocity of the body

  • Time taken by the body

SOLUTION :

Firstly lets calculate the time taken by the body through the second equations of motion

\underline{\boxed{\frak{\pm{\green{\ \: s = ut +  \dfrac{1}{2} at {}^{2}}}}}}\:\red{\bigstar}

s = 49

u = 0

a = 0.5

Substituting the values,

:\implies \sf \: 49 = 0(t) +  \dfrac{1}{2} (0.5)(t {}^{2} )

:\implies \sf \: 49 = 0 +  \dfrac{1}{2} ( \dfrac{1}{2} )(t {}^{2} )

:\implies \sf \: 49 = 0 +  \dfrac{1}{4}(t {}^{2} )

:\implies \sf \: 49 =   \dfrac{t {}^{2} }{4}

:\implies \sf \: 49  \times 4=   t {}^{2}

:\implies \sf \: t =  \sqrt{49 \times 4}

:\implies \sf \: t =  \sqrt{49 }  \times  \sqrt{4}

:\implies \sf \: t = 7 \times 2

 :\implies\bf \: t = 14s

So, time taken by the body is 14 s

From,

First equations of motion we find the final velocity of the body :-

\underline{\boxed{\frak{\pm{\green{\ \: v = u + at}}}}}\:\red{\bigstar}

u = 0

a = 0.5 m/s²

t = 14 s

 \sf:\implies \: v = 0 + (0.5)(14)

 \sf:\implies \: v = 0 + ( \dfrac{1}{2} )(14)

 \sf:\implies \: v = 0 + 7

 \bf :\implies\: v = 7m/s

So , final velocity of the body is 7 m/s

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