Question

A car initially at rest, picks up a velocity of 72 km/h In 1/4 minute. Calculate the acceleration and distance covered by the car.

Answer 1

Answer :-

• Acceleration = 1.3m/s^2

• Distance travelled = 146.25m

Explanation :-

Given :

• Initial velocity of the car,u = 0m/s

• Final velocity of the car,v = 72km/hr = 20m/s

• Time taken,t = 1/4 min = 15s

To Find :

• Acceleration,a = ?

• Distance,s = ?

Solution :

According to the first equation of motion:-

$\sf{}v=u+at$

$\implies\sf{}20=0+a\times 15$

$\implies\sf{}20=15a$

$\implies \sf{}\dfrac{20}{15}=a$

$\sf{}\therefore a=1.3m/s^2$

Therefore acceleration is equal to 1.3m/s^2

According to second equation of motion:-

$\sf{}s=ut+\dfrac{1}{2}at^2$

$\sf{}\implies s=0\times 15+\dfrac{1}{2}\times1.3\times15^2$

$\sf{}\implies s=0\times 15+\dfrac{1}{2}\times1.3\times225$

$\sf{}\implies s=0\times 15+146.25$

$\sf{}\implies s=0+146.25$

$\sf{}\therefore s=146.25m$

Therefore,distance travelled is 146.25m

Answer 2

Given :-

• A car initially at rest, picks up a velocity of 72 km/h in$\:\dfrac{1}{4}\:$minute

To Calculate :-

• Acceleration
• Distance covered by the car

Solution :-

We have,

• Initial Velocity(u) of the car = 0 m/s
• Final Velocity(v) of the car = 20 m/s
• Time taken(t) = $\:\dfrac{1}{4}\:$minute = 15 s

We know that,

$\boxed{\sf v = u + at} \: \: (\bf 1st \: equation \: of \: motion)$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time taken

Substituting the values we get,

$:\implies\sf20 = 0 + a \times 15$

$:\implies\sf20 = 15a$

$:\implies\sf\cancel- 15a = \cancel- 20$

$:\implies\sf a = \dfrac{\cancel{20}}{\cancel{15}}$

$:\implies\sf\underline{\boxed{\blue{\sf a = 1.3\:m/s^{2}}}}$

$\green{\therefore\sf Acceleration \: is \: 1.3\:m/s^{2}}$

We know that,

$\boxed{\sf s = ut + \dfrac{1}{2}at^{2}} \: \: (\bf 2nd \: equation \: of \: motion)$

Where,

• s = Distance Covered
• u = Initial Velocity
• t = Time taken

Substituting the values we get,

$:\implies\sf s = 0 \times 15 + \dfrac{1}{2} \times 1.3 \times 15^{2}$

$:\implies\sf s = 0 \times 15 + \dfrac{1}{2} \times 1.3 \times 225$

$:\implies\sf s = 0 \times 15 + \dfrac{1}{\cancel2} \times \cancel{292.5}$

$:\implies\sf s = 0 \times 15 + 146.25$

$:\implies\sf s = 0 + 146.25$

$:\implies\sf\underline{\boxed{\blue{\sf s = 146.25\:m}}}</p><p>$

$\green{\therefore\sf Distance \: travelled \: is \: 146.25\:m}$

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Additional Information :-

★ 3rd equation of motion :-

$\boxed{\sf v^{2} - u^{2} = 2as}$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance travelled

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