Question

A car goes from 4.47 m/s to 17.9 m/s in 3 seconds. Find the acceleration of the car.​

Answer 1

Answer:

4.476667 m/s^2

Explanation:

u=4.47m/s

v=17.9m/s

t=3sec

By using 1st equestion of motion,

v = u +at

17.9 = 4.47 + a(3)

17.9-4.47 = a(3)

a = 13.43 / 3

a = 4.47667 m/s^2

Answer 2

$\huge\blue{\tt Given :-}$

• A car goes from 4.47 m/s to 17.9 m/s in 3 seconds

$\huge\orange{\tt To \: Find :-}$

• Acceleration of the car

$\huge\green{\tt Solution :-}$

We have,

• Final Velocity(v) = 17.9 m/s
• Initial Velocity(u) = 4.47 m/s
• Time taken(t) = 3 s

We know that,

$\boxed{\tt v = u + at} \: \: [ \bf 1st \: equation \: of \: motion ]$

Where,

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

Substituting the give values we get,

$:\implies\tt 17.9 = 4.47 + a(3)$

$:\implies\tt 17.9 = 4.47 + 3a$

$:\implies\tt - 3a = 4.47 - 17.9$

$:\implies\tt \cancel- 3a = \cancel- 13.43$

$:\implies\tt a = \dfrac{\cancel{13.43}}{\cancel3}$

$:\implies\tt a = 4.4766666667$

$:\implies\orange{\tt a ≈ 4.48 \: m/s^{2}}$

$\therefore\underline{\bf Acceleration \: of \: the \: car \: is \: 4.48 \: m/s^{2}}$

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Additional Information :-

★ 2nd equation of motion :-

→ s = ut + ½ at²

Where,

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

★ 3rd equation of motion :-

→ v² - u² = 2as

Where,

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

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