Question

A train starting from rest attains a velocity of 72 km / h in 5 minutes. Assuming that the acceleration is uniform. find (i) the acceleration and (i) the distance traveled by the train for attaining this velocity -​

Answer 1

Answer:

$\underline{\purple {\ddot {\Mathsdude}}}$

☆Answered by Rohith kumar maths dude :-

☆Given that :-

▪Initial velocity=u=0m/s

▪Final velocity = 72km/h

▪Time =5minutes =5/60= 1/12hours .

☆To prove: -

▪i)acceleration

▪ii)Distance travelled

☆Proof:-

●We already knew that,

●Acceleration =a=v-u/t

a=72-0/1/12

a=72×12

a=864km/h^2

●Distance travelled =s=ut+1/2at^2

s=0×1/12+1/2×864×(1/12)^2

s=3km.

Hope it helps u mate.

Thank you

Answer 2

Answer:

$\huge\fcolorbox{red}{yellow}{ \color{cyan}Given}$

• Initial Velocity (U)=0,
• Velocity After some Time (v) =75 kmph

Converting into Meters

• 75×5/18 =≠=≠=>20 m/s

•Time (t)= 5 Min

5×60= 300 sec

Now, The Acceleration (a) = v-u/t

• 20-0/300 = 1/15 m/s

Distance Traveled for attaining that velocity is

• from v²-u²=2as

s=v²/2a (since u=0)

20²/2×(1/15)=3000 = 3km