Question

a train starting from rest attens q V of 72 meter per second square in 5 mint assuming uniform motjon accelerqtiob find the acceleration and the distqnce travelled by train of attaing its velocity​

Answer 1

Answer:

Acceleration of the train is$0.24 m/s^2$  and the distance travelled by train of  is 10.8 km

Explanation:

Time taken t=5 minute= 300 sec

Final speed of the train is 72 m/s

Initial velocity u=0 m/s

The acceleration o the train is

$v=u+at\\\\72=0+a \times 300\\\\a=0.24 m/s^2$

Distance traveled by train is

$S=ut+\frac{1}{2} at^2\\\\S=0 \times 120 +\frac{1}{2} \times 0.24 \times 300^2\\\\S=10800 m= 10.8 km$