Question

23. A bus increases its speed from 36 km/h to
72 km/h in 10 second then distance traveled by
bus during this time is
100 m
125 m
120 m
150 m​

Answer 1

★ Given :

bus changes its speed from 36km/h to 72 km/h in 10seconds.

u (initial velocity) = 36 km/h

v (final velocity) = 72 km/h

t = 10s

★ To find :

distance travelled by bus during this time.

★ Solution :

We know that,

★ v = u + at

★ 1 km/h = 5/18 m/s

→ 36 km /h = 10 m/s

36 × 5/18 =

5 × 2 = 10 m/s

→ 72 km/h = 20 m/s

72 × 5/18 =

4 × 5 = 20 m/s

by putting the values -:

20 = 10 + 10a

20 - 10 = 10 a

10 = 10 a

→ a = 1 m/s²

We know that ,

★ s = ut + 1/2 at²

by putting the values

= (10 × 10) + 1/2 × 1 × 10 × 10

= 100 + 1/2 × 100

= 100 + 50

= 150 m

$\purple{\therefore \: distance \: covered = 150 \: m}$

Answer 2

We know that,

v = u + at

1 km/h = 5/18 m/s

=>36 km /h = 10 m/s

36 × 5/18 =

5 × 2 = 10 m/s

=> 72 km/h = 20 m/s

72 × 5/18 =

4 × 5 = 20 m/s

Lets put he values -:

20 = 10 + 10a

20 - 10 = 10 a

10 = 10 a

=> a = 1 m/s²

As we know ,

s = ut + 1/2 at²

Lets put the values

= (10 × 10) + 1/2 × 1 × 10 × 10

= 100 + 1/2 × 100

= 100 + 50

= 150 m

$\blue{\therefore \: Distance \: Covered \: is \: 150 \: m}$