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 (ii) the distance
travelled by the train for attaining this
velocity​

Answer 1

Solution:

Given:

• Initial velocity(u) = 0 m/s
• Final Velocity(v) = 72 km/h

= 72 × 5/18

= 20 m/s

• Time = 5 min = 5 × 60 = 300 sec

To Find:

• Acceleration
• Distance travelled by train

Now, firstly we will calculate Acceleration,

=> Acceleration = (v - u)/t

=> Acceleration = (20 - 0)/300

=> Acceleration = 20/300

=> Acceleration = 1/15 m/s²

Hence, Acceleration = 1/15 m/s²

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Now, we will calculate distance travelled by train by 2nd equation of motion,

=> s = ut + 1/2 at²

=> s = 0 + 1/2 × 1/15 × (300)²

=> s = 0 + 1/2 × 1/15 × 90000

=> s = 1/2 × 6000

=> s = 3000 m

Now, we know that, 1 km = 1000 m

So, 3000 m = 3 km

Hence, Distance travelled by train = 3 km.

Answer 2

Answer:

Given :

u=0m/s

V= 72km/h

T =5 min =5/60 h =1/12 h

To find:

a=?

and distance

we know  , a=v-u/t

⇒a=72-0/1/12

=72*12

=864km/h²

s=ut+1/2at²

=0x1/12 +1/2(864)(1/12)²

=(1/2)x864x(1/144)

=3km

∴Distance travelled by the train  3km.