Physics, asked by bishal5006, 8 months ago

 A car starting from rest acquires a velocity of 54 km/h in 5 seconds. calculate its acceleration and displacement .


Answers

Answered by Anonymous
20

Answer :

➥ Acceleration of the car = 7.5 m/s²

➥ Displacement of the car = 15 m

Given :

➤ Intial velocity of the car (u) = 0 m/s

➤ Final velocity of the car (v) = 54 km/h

➤ Time taken (t) = 2 sec

To Find :

➤ Acceleration of the car (a) = ?

➤ Displacement of the car = ?

Solution :

◈ Final velocity (v) = 54 km = 54 × ⁵/₁₈ = 15 m/s

Acceleration is given by

 \tt{: \implies a =  \dfrac{v - u}{t} }

 \tt{: \implies a =  \dfrac{15 - 0}{2} }

 \tt{: \implies a =   \cancel{\dfrac{15}{2} }}

  \tt{: \implies  \green{ \underline{ \overline{ \boxed{ \purple{ \bf{ \:  \: a = 7.5 \: m/s^2 \:  \: }}}}}}}

Hence, the acceleration of the car is 7.5 m/s².

Now ,

We calculate displacement of the car

From second equation of motion

 \tt{: \implies s = ut +  \dfrac{1}{2} a {t}^{2}  }

 \tt{: \implies s = 0 \times 2 +  \dfrac{1}{2} + 7.5 \times 2 \times 2 }

 \tt{: \implies 0 +  \dfrac{1}{ \cancel{2}} \times 7.5 \times  \cancel{2} \times 2 }

 \tt{: \implies s = 0 + 1 \times 7.5 \times 2}

 \tt{: \implies s = 0 + 15 }

 \tt{: \implies s =  15}

  \tt{: \implies  \green{ \underline{ \overline{ \boxed{ \purple{ \bf{ \:  \: s = 15 \: m \:  \: }}}}}}}

Hence, the displacement of the car is 15 m.

Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

Answered by Anonymous
5

\small{\underline{\sf{\red{Given:—}}}}

☛︎ Initial velocity of the car, u = 0 (at the rest)

☛︎ Final velocity of the car, v = 54 km/h = 15m/s

☛︎ Tme taken, t = 5 seconds

\small{\underline{\sf{\red{To\: find:—}}}}

  • Acceleration of the car, a = ?
  • Displacement of the car, s = ?

\small{\underline{\sf{\red{Solution:—}}}}

\boxed{\bf{\blue{Acceleration, a=\frac{v-u}{t}}}}

➪Acceleration\:of\:the\:car,\:a=\frac{15–0}{5}m/s²

\boxed{\bf{\green{Acceleration,\:a=3m/s²}}}

NOW,

\boxed{\bf{\blue{Displacement,\:s=ut+\frac{1}{2}at²}}}

➪Displacement=0×5+\frac{1}{2}×3×5² m

➪Displacement,\:s=0+\frac{1}{2}×75 m

\boxed{\bf{\green{Displacement,\:s=37.5\:m}}}

Therefore, Acceleration of the car is 3m/ and

Displacement of the car is 37.5 m

__________________________

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