Question

A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds calculate (i) acceleration in m/s? 2 ml? (ii) distance covered by the bus in metres during this interval.

Answer 1

Answer:

acceleration=.5m/s2

distance covered=175meter

Answer 2

Given:-

• Uniform velocity of bus, u = 54 km/h.
• Final velocity of bus, v = 72 km/h.
• Time taken, t =10 seconds.

To Find:-

• Acceleration of bus is m/s ?
• Distance covered by bus ?

Formula to be used:-

• v = u + at
• s = ut + ½ at²

Solution:

$\qquad$☀️First we need to change the uniform and final velocity into meter/second. To change multiply by 5/18.

$\qquad$❏ Uniform velocity (u) = 54(5/18)

$\qquad$$:\implies$u = 15 m/s

$\qquad$❏ Final velocity (v) = 72(5/18)

$\qquad$ $:\implies$v = 20 m/s.

Using First equation of motion :-

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ v = u + at}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{20 = 15 + a(10)}}$

$\qquad\leadsto\quad \pmb {\mathfrak{20 – 15 = 10\times a}}$

$\qquad\leadsto\quad \pmb {\mathfrak{5 = 10\times a}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ \cancel{\dfrac{5}{10}} = a}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{a= 0.5 m/s²}}}$

• Acceleration of the bus is 0.5 m/s².

Using Second equation of motion :-

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{s = ut + ½ at²}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{s = 15 \times 10 + ½ \times 0.5 \times (10)²}}$

$\qquad\leadsto\quad \pmb {\mathfrak{s = 150 + \dfrac{0.5}{2}\times 100}}$

$\qquad\leadsto\quad \pmb {\mathfrak{s = 150 + 0.5(50)}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ s = 150 + 25}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ s = 175 m}}}$

• Hence, the distance covered by bus is 175 m.

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$\qquad$☀️Equations of motion :-

$\qquad$ ⑴ v = u + at

$\qquad$ ⑵ s = ut + ½at²

$\qquad$ ⑶ v² - u² = 2as

• v = final velocity
• u = initial velocity
• s = displacement
• t = time taken
• a = acceleration

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