Question

a bus starting from rest moves with a uniform acceleration of 0.6m/s for 6 minutes find the speed acquired and distance travelled ​

Answer 1

GiveN :

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 0.6 m/s²
• Time interval (t) = 6 mins

To FinD

• Speed Acquired (Final velocity)
• Distance Travelled

SolutioN :

Convert Time into seconds, by multiplying it with 60.

$\implies \sf{Time = 6 mins} \\ \\ \\ \implies \sf{Time = 6 \times 60} \\ \\ \\ \implies \sf{Time = 360 s}$

___________________

Use 1st equation of Kinematics :

$\Large{\boxed{\sf{v = u + at}}} \\ \\ \\ \implies \rm{v = 0 + 0.6 \times 360} \\ \\ \\ \implies \rm{v = 216 ms^{-1}} \\ \\ \\ \underline{\rm{\therefore \: Final \: velocity \: is \: 216 \: ms^{-1}}}$

____________________

Now, use 3rd equation of Kinematics :

$\Large \boxed{\sf{v^2 - u^2 = 2as}} \\ \\ \\ \implies \rm{(216)^2 - 0^2 = 2 \times 0.6 \times s} \\ \\ \\ \implies \rm{4656 - 0 = 1.2s} \\ \\ \\ \implies \rm{s = \dfrac{4656}{1.2}} \\ \\ \\ \implies \rm{s = 38880 m} \\ \\ \\ \implies \rm{s = 38.88 km} \\ \\ \\ \underline{\rm{\therefore{Distance\ travelled\ is\ 38.88 km}}}$

Answer 2

Given :-

Initial speed of the bus = 0 m/s

Acceleration = 0.6 m/s²

Time taken, t = 6 minutes = 360 s

To Find :-

(a) Speed acquired,

(b) Distance travelled.

Solution :-

Putting all the value, we get

(a) Speed acquired

⇒ v = u + at

⇒ v = 0 + 0 × 0.6 × 360

⇒ v = 216 m/s

(b) Distance travelled.

⇒ (12)² - (0)² = 2(0.1) s

⇒ s = 720 m

Hence, Speed acquired finally by the bus is 12 m/s.

And Distance travelled by the bus is 720 m.