Question

A bus accelerates uniformly from
15m/s to 20m/s in 10 seconds. Calculate
the distance covered by the bus during
this time interval.​

Answer 1

Given :

• Initial velocity of bus, u = 15 m/s
• Final velocity of bus, v = 20 m/s
• time taken to attain final velocity, t = 10 s

To find :

• Distance travelled by bus during this time interval, s =?

Formula required :

• First equation of motion

      v = u + a t

• Second equation of motion

     s = u t + 1/2 a t²

[ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken, s is distance covered ]

Solution :

Using the first equation of motion

→ v = u + a t

→ 20 = 15 + a ( 10 )

→ 20 - 15 = 10 a

→ 5 = 10 a

→ a = 5/10

→ a = 0.5 m/s²

Using the second equation of motion

→ s = u t + 1/2 a t²

→ s = (15) (10) + 1/2 (0.5) (10)²

→ s = 150 + 25

→ s = 175 m

Therefore,

• Distance covered by Bus is 175 m.

Answer 2

$\star\:\:\:\bf\large\underline\green{Given:—}$

☛︎ Initial velocity (u)= 15 m/s

☛︎ Final velocity (v) = 20 m/s

☛︎ Time taken (t) = 10 seconds

$\star\:\:\:\bf\large\underline\green{To\:find:—}$

Distance covered by bus during this time interval (s) = ?

$\star\:\:\:\bf\large\underline\green{Solution:—}$

We know that,

$\boxed{\bf{\red{v=u+at}}}$

↦20 = 15 + a × 10 m/s²

↦20 - 15 = a×10 m/s²

↦5 = 10a m/s²

↦10a = 5 m/s²

↦a = 0.5 m/s²

Acceleration of the 0.5 m/s²

NOW,

As per 2nd equation of motion :-

$\boxed{\bf{\red{S=ut+\frac{1}{2}at²}}}$

↦$S=15×10+\frac{1}{2}×0.5×(10)²\:m$

↦$S=150+\frac{1}{2}×0.5×100\:m$

↦$S=150+25\:m$

$\boxed{\bf{\purple{S=175\:m}}}$

Hence,the distance covered by the bus during this time interval is 175 m

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