Question

A train accelerates from 36 km/h to 72 km/h in 20 s. Find the distance travelled.

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Answer 1

Explanation:

36 km/hr=36*5/18=10 m/sec

72 km/hr=72*5/18=20 m/sec

t=20 sec

v=u+at

20=10+20a

a=1/2 m/(sec)2

s=ut+1/2at2

s=(10*20)+1/2*1/2*20*20

s=200+100

s=300m

Answer 2

Answer :-

Given :-

• Initial velocity = 36 km/hr = 36 × 5/18 = 10m/s
• Final velocity = 72 km/hr = 72 × 5/18 = 20 m/s
• Time taken = 20 sec

To Find :-

• Distance travelled

Solution :-

To calculate distance travelled, we first need to find acceleration -

Calculating Acceleration -

• u = 10 m/s
• v = 20 m/s
• t = 20 sec

Substituting the value in formula :-

➩ $\rm a = \dfrac{v-u}{t}$

➩ $\rm a = \dfrac{20-10}{20}$

➩ $\rm a = \dfrac{1 \cancel 0}{2\cancel 0}$

➩ $\rm a = \dfrac{1}{2}$

$\rm Acceleration = 0.5 m/s^2$

Calculating Distance travelled -

• a = 0.5 m/s²
• u = 10 m/s
• t = 20 sec

Substituting the value in 2nd equation of motion -

➩ $\rm s = ut + \frac{1}{2} at^2$

➩ $\rm s = 10 \times 20 + \frac{1}{2} \times 0.5 \times 20^2$

➩ $\rm s = 200 + 100$

➩ $\rm s = 300 m$

Distance travelled = 300 m