Question

A train starting from rest with an 70.5m s^-2 to for 2 min. calculate the valocity aquirrd and distance travelled by train

Answer 1

Answer:

8640 m/s

507.6 km

Explanation:

Given:

Initial velocity = u = 0m/s

Acceleration = a = 70.5 m/s²

time = t= 2 minutes = 2×60 seconds = 120 seconds

To find:

Final velocity (v)

Distance (s)

We can get the final velocity of the train by applying the formula : $V=u+at$

So substituting the values we get:

$V=0+70.5 \times 120$

$V=0+8460$

$V=8460\ m/s$

The final velocity of the train is equal to 8,460 m/s

Now we can use the third equation of motion to get the distance which says: $v^{2} -u^{2} =2as$

Substituting the values we get:

$8460^{2} -0^{2} =2 \times 70.5 \times s$

$= 71571600-0 = 141 \times s$

$71571600 = 141s$

$s=\frac{71571600}{141}$

s= 5,07,600 m

1 km =1000m

5,07,600 = 507.6 km

$507.6 km$

The final velocity is equal to 8640 m/s and distance covered is equal to 507.6 km

Answer 2

Given :-

Initial velocity (u)=0

Time (t)= 2×60 =120sec

Acceleration (a)=70.5m/s^2

To find :-

Final velocity (v) and Distance (s)

• Now by 1st equation of motion

$\bigstar{\boxed{\sf{v=u+at}}}$

$\dashrightarrow\sf v=0+70.5\times 120\\ \\ \dashrightarrow\sf v= 0+8460\\ \\ \dashrightarrow\sf v= 8460m/s$

$\underline{\sf{\dag \ \ Final\ velocity (v)= 8460m/s}}$

• Now the Distance covered !
• By applying 2nd Equation of motion we can get the distance

$\bigstar{\boxed{\sf{s=ut+\dfrac{1}{2}at^2}}}$

• We have

$\longmapsto\sf acceleration (a)=70.5m/s^2\\ \\ \longmapsto\sf final \ velocity (v)= 8460m/s\\ \\ \longmapsto\sf initial \ velocity (u)=0\\ \\ \longmapsto\sf Time (t)= 120 \ second$

• Put the given values

$\dashrightarrow\sf s=0\times 120+\dfrac{1}{\cancel2}\times \cancel{70.5} \times (120)^2\\ \\ \dashrightarrow\sf s= 0+35.25\times 14400\\ \\ \dashrightarrow\sf s=507,600m$

• Now convert it into Kilometer

$\bullet\sf 1 km= 1000m\\ \\\dashrightarrow\sf \dfrac{507600}{1000}km\\ \\ \dashrightarrow\sf 507.6km$

$\boxed{\sf{\dag \ Final \ velocity (v)= 8460m/s}}$

$\boxed{\sf{\dag \ Distance (s) =507.6km}}$