Question

A bus moving with a velocity 6m/s stopped by applying breakes after 3m,acceleration =?​

Answer 1

Answer:

Acceleration = -6m/s^2

Explanation:

Initial velocity, u = 6m/s

Final Velocity, v = 0m/s

Distance covered, s = 3m

We know that, v^2 - u^2 = 2as

=> 0 - 36 = 6a

=> a = -6m/s^2

Hope this helps.....

Answer 2

$\huge\underline {\sf {Answer :}}$

$\underline {\rm{Given :}}$

• A bus moving with a velocity 6m/s.
• It stopped by applying breakes after 3m.

$\underline {\rm{Find :}}$

• The acceleration = ?

$\underline {\rm{Solution :}}$

$\dag$Formula to be used :

$\boxed {\bold {{v}^{2} = {u}^{2} + 2as }}$

$\bigstar$ From the Question we have :-

• Initial velocity (u) = 6m/s
• Final velocity (v) = 0
• Distance (s) = 3m
• Acceleration (a) = ?

Do remind :

It is the third equation of motion.

It gives the velocity acquired by a body in travelling a distance s.

✬Now substituting the values in the formula :

= ${v}^{2} = {u}^{2} + 2as$

= ${0}^{2} = {6}^{2} + 2×a×3$

= ${0}^{2} - {6}^{2} = 2×a×3$

= $-36 = 6a$

= $\dfrac {-36}{6} = a$

= $-6 = a$

Thus, from the third equation of motion the acceleration of the bus is -6m/s.

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