Question

a train accelerates from 36 km/h to 54 km/h in 10 second. calculate. a–acceleration of train. b–distance travelled by train in 10 second .

Answer 1

First we need to convert the velocities given from km/h to m/s
It comes out to be
u=10 m/s (36km/h)
v= 15m/s (54km/h)

We know,
v=u+at
=> 15 = 10+ a × 10
=> 10 ×a = 5
=> a= 0.5 m/ s2

Again ,
v2= u2+ 2as
=>15^2= 10^2+ 2×0.5 ×s
=> s= 125m.


Hope you understand...

Answer 2

HOLA! !!

HERE YOUR ANSWER IS

GIVEN

Initial velocity (u) = 36km/hr = 36000m /3600s


=10m/s


final velocity (v) = 54km/hr = 54000/3600m/s

= 15m/s

acceleration =?

distance = ?

SOLUTION


Acceleration = (v-u)/t

= (15-10)/10

= 5/10 =1/2ms^-2

acceleration = 0.5ms^-2

We know a kinematical equation


s = ut+1/2at^2

= 10 (10) + 1/2 (0.5)(10)^2

= 100+1/2 (0.5)(100)

= 100 +1/2 (50)

= 100 + 25

s = 125m

Therefore a = 0.5ms^-2

s =125m


HOPE IT HELPS YOU :D