Question

A bus accelerate uniformly from 54km/h to 74km/h in 10s. Calculate 1.acceleration
2.distance travelled​

Answer 1

Answer:

Step-by-step explanation:

Given :-

Initial velocity, u = 72 km/h = 72 (5/18) m/s = 20 m/s  

Final velocity, v = 54 km/h = 54 (5/18) m/s = 15 m/s  

Time in which velocity changes is, t = 10 seconds  

To Find :-

(a) Acceleration

(b) Distance traveled

Formula to be used :-

a = (v - u)/t  and v² = u² + 2aS  

Solution :-

(a) Acceleration

Acceleration, a = (v - u)/t  

⇒ a = (20-15)/10

= 0.5 m/s²

(b) Distance traveled by the bus  

v² = u² + 2aS  

⇒ 20² = 15² + 2 × 0.5 × S  

⇒ 400 = 225 + 1×S

⇒ 400 - 225 = S

⇒ S = 175 m

Hence, the acceleration and distance traveled​ by bus is 0.5 m/s² and 175 m.

Answer 2

$\large \bold{ \underline{ \underline{ \sf \: Answer : \: \: \: }}}$

$\star$ Acceleration = 0.5 m/s²

$\star$ Distance = 175 m

$\large \bold{ \underline{ \underline{ \sf \: Explaination : \: \: \: }}}$

(i) Acceleration

$\large \bold{ \fbox{ \fbox{ \sf Acceleration = \frac{(v - u)}{t} }}}$

$\sf \to a = \frac{ (20-15)}{10} </p><p> \\ \\ </p><p> \sf \to a = 0.5 \: \: m/{s}^{2}$

(ii) Distance traveled by the bus

$\large \fbox{ \fbox{ \bold{ \sf {v}^{2} = {u}^{2} + 2aS }}}$

$\to \sf {(20)}^{2} = {(15)}^{2} + 2 × 0.5 × S \\ \\ \sf</p><p> \to</p><p>400 = 225 + 1×S \\ \\ \sf </p><p> \to</p><p>400 - 225 = S \\ \\ \sf</p><p> \to</p><p>S = 175 \: m$

$\therefore$ The acceleration and distance traveled by bus is 0.5 m/s² and 175 m