Question

a train starting from rest, attains a velocity of 36 km/hr in 4 sec. Find its

A) acceleration
B) distance travelled in 4 sec
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Answer 1

Answer:

A)2.5 m/$s^{2}$

B)20 m

Explanation:

Initial velocity (u) = 0 m/s

Final velocity (v) = 36 km/hr=10 m/s

Time taken (t) = 4 s

Therefore,

Acceleration = (v-u)/t

                     = (10-0)/4

                     = 10/4

                     = 2.5 m/$s^{2}$

Now, let distance travelled in 4 sec be S m.

Therefore, from the second equation of motion, we know that:

S=ut+$\frac{at^{2} }{2}$

=>S=0+(2.5*$4^{2}$)/2

=>S=(2.5*16)/2

=>S=2.5*(16/2)

=>S=2.5*8

=>S=20 m