Question

A bus starting from the rest attains a velocity of 90km/hr in 10 sec. acceleration and distance covered by the car is*
3 points.

i) 250m/s & 125m

ii) 2.5cm/s^2 & 125 m

iii)2.5m/s & 125km

iv) 2.5m/s^2 & 125m

​

Answer 1

Given :

• Bus started from rest has attained a velocity of 90 km/hr in 10 sec

To find :

• The acceleration and distance covered by the bus

Solution :

We have ,

• u = 0 m/s (started from rest)
• v = 90 km/hr = 25 m/s
• t = 10 sec

From first equation of motion ,

$\displaystyle \: \dag \boxed {\rm{v = u + at \: }}$

Here ,

• v is final velocity
• u is initial velocity
• a is acceleration
• t is time

By substituting the values ,

$: \implies \rm \: 25 \: m {s}^{ - 1} = 0 + a(10 \: s) \\ \\ : \implies \rm \: 25 \: m {s}^{ - 1} = a(10 \: s) \\ \\ : \implies \rm \: \frac{25 \: m {s}^{ - 1} }{10 \: s} = a \\ \\ : \implies \rm \: a = 2.5 \: m {s}^{ - 2}$

Hence , The acceleration of the bus is 2.5 m/s²

Now , From second equation of motion

$\displaystyle \dag \boxed {\rm{ \: v {}^{2} - {u}^{2} = 2as \: }}$

Here ,

• v is final velocity
• u is intial velocity
• a is acceleration
• s is distance

By substituting the values ,

$: \implies \rm \: ( 25 \: m {s}^{ - 1} ){}^{2} - ( 0){}^{2} = 2(2.5 \: m {s}^{ - 2} )(s) \\ \\ : \implies \rm \: (625 \cancel{m {}^{2} {s}^{ - 2}} ) - 0 = (5 \cancel{ m {s}^{ - 2}} )s \\ \\ : \implies \rm \:625 \: m= 5s \\ \\ : \implies \rm \: \frac{625 \: m}{5} = s \\ \\ : \implies \rm \: 125 \: m = s \\ \\ : \implies \rm \: s = 125 \: m$

Hence , The distance covered by the bus is 125 m

∴ Option(4) is the required answer .

Answer 2

Explanation:

Given :

Bus started from rest has attained a velocity of 90 km/hr in 10 sec

To find :

The acceleration and distance covered by the bus

Solution :

We have ,

u = 0 m/s (started from rest)

v = 90 km/hr = 25 m/s

t = 10 sec

From first equation of motion ,

\displaystyle \: \dag \boxed {\rm{v = u + at \: }}†

v=u+at

Here ,

v is final velocity

u is initial velocity

a is acceleration

t is time

By substituting the values ,

\begin{gathered} : \implies \rm \: 25 \: m {s}^{ - 1} = 0 + a(10 \: s) \\ \\ : \implies \rm \: 25 \: m {s}^{ - 1} = a(10 \: s) \\ \\ : \implies \rm \: \frac{25 \: m {s}^{ - 1} }{10 \: s} = a \\ \\ : \implies \rm \: a = 2.5 \: m {s}^{ - 2} \end{gathered}

:⟹25ms

−1

=0+a(10s)

:⟹25ms

−1

=a(10s)

:⟹

10s

25ms

−1

=a

:⟹a=2.5ms

−2

Hence , The acceleration of the bus is 2.5 m/s²

Now , From second equation of motion

\displaystyle \dag \boxed {\rm{ \: v {}^{2} - {u}^{2} = 2as \: }}†

v

2

−u

2

=2as

Here ,

v is final velocity

u is intial velocity

a is acceleration

s is distance

By substituting the values ,

\begin{gathered} : \implies \rm \: ( 25 \: m {s}^{ - 1} ){}^{2} - ( 0){}^{2} = 2(2.5 \: m {s}^{ - 2} )(s) \\ \\ : \implies \rm \: (625 \cancel{m {}^{2} {s}^{ - 2}} ) - 0 = (5 \cancel{ m {s}^{ - 2}} )s \\ \\ : \implies \rm \:625 \: m= 5s \\ \\ : \implies \rm \: \frac{625 \: m}{5} = s \\ \\ : \implies \rm \: 125 \: m = s \\ \\ : \implies \rm \: s = 125 \: m\end{gathered}

:⟹(25ms

−1

)

2

−(0)

2

=2(2.5ms

−2

)(s)

:⟹(625

m

2

s

−2

)−0=(5

ms

−2

)s

:⟹625m=5s

:⟹

5

625m

=s

:⟹125m=s

:⟹s=125m

Hence , The distance covered by the bus is 125 m

∴ Option(4) is the required answer .