Question

a train starting from rest travelling with uniform acceleration and acquire a velocity of 80 kilometre per hour in 2 minutes find the acceleration and the distance travelled by the train during this time​

Answer 1

Answer: u= 0m/s

v= 80 km/h= 80/1*1000/3600= 22.22 m/s

t= 2 min= 60s * 2= 120s

by using the first  equation of motion i.e. (i) v= u + at

22.22 = 0 + a* 120

22.22 = a*120

22.22/120 = a

0.185167 m/s^2 =a

now we will use the third equation of motion i.e. (iii) 2as= v^2 - u^2

2*0.185167*s= (22.22)^2 -(0)^2

0.370334*s=493.72

s=493.72/0.370334

s= 1333.17 m

Explanation:

Answer 2

Given

⟡ Train starting from rest.

⟡ Final velocity = 80 km/h

⟡ Time taken = 2 minutes.

To find

◆ Acceleration & distance travelled by train.

Solution

As the train starts from rest, so initial velocity (u) = 0

Here, final velocity = 80 km/h

→ Final velocity (v) = 80 × (5/18) m/s

→ Final velocity (v) = 22.22 m/s

Again, time = 2 minutes.

→ Time = (2 × 60) = 120 s

$\underline{\dag{\textit{Using 1st equation of motion : }}} \\\\\longrightarrow\underline{\large{\boxed{\sf\green{v = u + at }}}}\: \bigstar \\\\\longmapsto\sf 22.22 = 0 + a \times 120 \\\\\longmapsto\sf 22.22 = 120a \\\\\longmapsto\sf a = \dfrac{22.22}{120}\\\\\longmapsto\underline{\boxed{\textsf{\textbf{a = 0.185 m/s ^{\text{2}} }}}}\\\\\\\therefore\underline{\textsf{Acceleration of train = {\textbf{0.185 m/s ^{\text{2}} }}}}$

$\rule{150}{1}$

Again,

$\underline{\dag{\textit{Using 3rd equation of motion : }}}\\\\\longmapsto\underline{\large{\boxed{\sf\blue{v^2 - u^2 = 2as }}}}\: \bigstar \\\\\longmapsto\sf (22.22)^2 - 0^2 = 2 \times 0.185 \times s \\\\\longmapsto\sf 493.7284 = 0.37s \\\\\longmapsto\sf s = \dfrac{493.7284}{0.37} \\\\\longmapsto\underline{\boxed{\textsf{\textbf{s = 1334 m }}}}\\\\\\\therefore\underline{\textsf{Distance travelled by train = {\textbf{1334 m }}}}$