Question

If the velocity of a train starting from rest becomes 72 km/h in 10 min .
(a) what is it's acceleration ?

(b). calculate the distance travelled by the train within this time interval
​

Answer 1

Answer:

Explanation:

u = 0

v = 72 km/h = 20 m/s

t = 10 min = 600 sec

(a)      a= v-u/t = (20-0)/600

                       ≈ 0.033 m/$s^{2}$

(b)    s = ut + (1/2)a$t^{2}$  

          = (1/2)*0.033*600*600

         = 5940 m

Answer 2

Given :

Velocity of the train from rest, V = 72 km/hr.

Times (t) = 10 min = 10/60 h = 1/6 h.

To find :

Acceleration = ?

Distance travelled bt the train = ?

Solution :

Acceleration, a = ?

a = v-u/t

a = 72-0/10

a = 72×10

a = 720 km/h².

Distance travelled = S = ?

S = ut+1/2at²

S = 0×1/10+12(720)(1/10)²

S = (1/2)×720×(1/100)

S = 360/100

S = 3.6 km.

•°• The acceleration is 720 km/hr .

•°• The distance travelled by the train within this travel time is 3.6 km.