1.

А іA car is moving at 72 km/her after covering adistance of 100m his nelocity is 54 km/hr calculatethe acceleration of the car ?​

Answer»

\huge {\bold {\underline {\underline{Given : - }}}}

  • Initial VELOCITY of car = 72km/hr
  • Distance covered = 100m
  • Final velocity = 54 km/hr

\huge {\bold {\underline {\underline{To \: find : - }}}}

  • Acceleration of car

\huge {\bold {\underline {\underline{Solution : - }}}}

The relation between km/hr and m/s is given by ,

\large {\underline {\bold {\boxed{ \bigstar{ | 1km.hr {}^{ - 1} = \dfrac{5}{18} \: ms {}^{ - 1} }}}}}

We are given that ,

  • \small\rm{v = 54km.hr {}^{ - 1} = \frac{54 \times 5}{18} = 15 \: m. {s}^{ - 1}}

  • \small\rm{ u = 72km. {hr}^{ - 1} = \dfrac{72 \times 5}{18} = 20 \: m. {s}^{ - 1} }

  • S = 100m

From THIRD equation of MOTION ,

\large \underline{\bold {\boxed {\bigstar{ | {v}^{2} - {u}^{2} = 2as}}}}

where ,

  • v is final velocity
  • u is initial velocity
  • a is acceleration
  • s is distance

: \implies \: (15) {}^{2} - (20) {}^{2} = 2(a)(100) \\ \\ : \implies \: 225- 400 = 200a \\ \\ : \implies \: - 175 = 200a \\ \\ : \implies \: a = - \dfrac{ 175}{200} \\ \\ : \implies \: a = - 0.875 \: m. {s}^{ - 2}

∴ The Acceleration of the car is -0.875 m/s²

\huge {\bold {\underline {\underline{Additional \: info: - }}}}

★ The first equation of motion is given by ,

\large \underline {\bold {\boxed {\bigstar{ \: | v = u + at}}}}

Where ,

  • v is final velocity
  • u is initial velocity
  • a is acceleration
  • t is time

★ Second equation of motion is given by ,

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

Where ,

  • S is distance
  • u is initial velocity
  • a is acceleration
  • t is time

\huge {\bold {\underline{\underline {Note : - }}}}

Equations of Motion are valid only in case of constant acceleration.If acceleration has negative sign it indicates the DECREASE in acceleration (or) deceleration.


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