Question

A bus starting from rest acquires a velocity of 15 m/s in 30 s. Calculate acceleration and distance traveled by the bus during this time. (2)

Answer 1

${\huge{\bf{\red{\underline{\underline{Given :}}}}}}$

A bus starting from rest acquires a velocity of 15 m/s in 30 s.

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${\huge{\bf{\blue{\underline{\underline{To Find:}}}}}}$

Calculate acceleration and distance traveled by the bus during this time.

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${\huge{\bf{\orange{\underline{\underline{Formula \: used :}}}}}}$

${\bf{\pink{A = \frac{v - u}{t} }}}$

${\bf{\pink{Distance = speed \times time}}}$

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${\huge{\bf{\green{\underline{\underline{Solution :}}}}}}$

U = 0m/s as the bus was on rest

V = 15m/s

T = 30s

${\bf{A = \frac{v - u}{t} }}$

${\bf{A =}}{\bf{\cancel{ \frac{0 - 15}{30} }}}$

${\sf{\red{A = 2m/s}}}$

Distance:

${\sf{Distance = speed \times time}}$

${\sf{Distance = 15 \times 30}}$

${\sf{\red{Distance = 450 m}}}$

So,

${\mathfrak{\orange{The \: acceleration = 2m/s \: and \: distance = 450m}}}$

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${\huge{\blue{\boxed{\boxed{XitzcuteChandanX}}}}}$

Answer 2

Good Morning My dear friend, here is the answer to your question,

Here,

u = 0 m/s

v = 15 m/s

t = 30 sec

we know that,

v = u + at --------------------------( first equation of motion)

Therefore,

15 = 0 + a × 30

a = ½ m/s^2

v^2=u^2+2as--------------------( third equation of motion )

Therefore,

15^2= 0^2+2 ×1/2×s

225 m =s