Question

8.1) A train accelerates from 36 Km/h to 54km/h in 10
sec
i) Acceleration
ii) The distance travelled
by car.​

Answer 1

Answer:

Given:

Initial velocity= 36km/h=36x5/18=10m/s Final velocity =54km/h3D54x5/18=15m/s Time =10 sec

Acceleration = v-u/ t =

=15-10/10=5/10=1/2=0.5 m/s2 Distance =s=?

From second equation of motion:

S=ut +1/2 at^2 =10*10+1/2*0.5*10*10 =100+25 =125m So distance travelled is 125m

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Answer 2

Answer:

Acceleration = 0.5 m/s²

Distance travelled = 125 metres

Explanation:

Given :

• Initial velocity = u = 36 km/hr

• Final velocity = v = 54 km/hr

• Time taken = t = 10 seconds

To find :

• Acceleration of the train

• Distance travelled by the train

36 km/hr = 36×5/18 = 10 m/s

54 km/hr = 54×5/18 = 15 m/s

Acceleration = (15-10)/10

Acceleration = 5/10

Acceleration = 0.5 m/s²

Using the third equation of motion :

V²-u²=2as

15²-10²=2×0.5×s

225-100 = 1s

125 = s

The acceleration of the train is 0.5 m/s² and distance travelled by it is 125 metres