Question

A bus which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s² over a distance of 340 m. How fast is it going after that acceleration? ​

Answer 1

Answer:

Formulae,

ut+21at2=50t

0+21×5×t2=50t

t=50×52=20s

v2=u2+2as

v2=0+2×5×(50×20)

v2=10000

∴v=100m/s

Answer 2

Given :

• Intial Velocity = 15 m/s.

• Accelaration = 6.5 m/s².

• Distance covered = 340 m.

To find :

The final Velocity of the bus.

Solution :

We know the third Equation of Motion i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² ± 2aS

Where :

• v = Final Velocity.
• u = Initial Velocity.
• a = Acceleration.
• S = Distance traveled.

Using the third Equation of Motion and substituting the values in it, we get :

==> v² = u² ± 2aS

[Note :- Since the body is Accelarating the Acceleration Produced by the bus will be postive]

==> v² = u² + 2aS

==> v² = 15² + 2 × 6.5 × 340

==> v² = 225 + 2 × 6.5 × 340

==> v² = 225 + 2 × 65/10 × 340

==> v² = 225 + 2 × 65 × 34

==> v² = 225 + 4420

==> v² = 4645

==> v = √4645

==> v = 68.15

∴ v = 68.15 m/s.

Hence the final Velocity of the bus is 68.15 m/s.