Question

The speed of a bus increases uniformly from 10m/s to 50m/s in 12s. calculate the distance travelled during the entire period.

Answer 1

Answer:-

Given:

Initial Velocity (u) = 10 m/s

Final velocity (v) = 50 m/s

Time (t) = 12 s.

We know that,

v = u + at

[ a is the acceleration ]

So,

⟶ 50 = 10 + (a) (12)

⟶ 50 - 10 = 12a

⟶ 40/12 = a

⟶ 3.33 m/s² = a

Now,

using second equation of motion,

⟶ S = ut + (1/2) * at²

⟶ S = (10)(12) + (1/2)(3.33)(12)²

⟶ S = 120 + 239.76

⟶ S = 359.76 m

∴ The distance travelled by the bus is 359.76 m.

Answer 2

Given :

• The speed of a bus increases uniformly from 10m/s to 50m/s in 12s.

To Find :

• Calculate the distance travelled during the entire period

Solution :

Concept :

• The second equation of motion gives the displacement of an object under constant acceleration:

$: \implies \: \: \: \: \: \: \boxed{ \sf \:v = u + at}$

Substitute all values :

$: \implies \: \: \: \: \: \: \sf \:50= 10+ 12a \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:50 - 10 = 12a \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:40 = 12a \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:a = \cancel{ \frac{40}{12}}\\ \\ \\ : \implies \: \: \: \: \: \: \sf \:a = 3.33$

$\\ \\ : \implies \: \: \: \: \: \: \boxed{ \sf \:s= ut + \frac{1}{2} a {t}^{2} }$

Substitute all values :

$\\ \\ : \implies \: \: \: \: \: \: \sf \:s= 10 \times 12+ \frac{1}{2} \times 3.33 \times {12}^{2} \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:s= 120 + \frac{1}{ \cancel{2}} \times 3.33 \times \cancel{ 144} \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:s= 120 + 239.76 \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:s= 359.76$

The distance travelled by the bus is 359.76 m.