Question

FORMULA FOR FINDING DISTANCE S = UT+½AT²

A TRAIN STARTING FROM REST MOVES WITH UNIFORM ACCELERATION OF 0.2 M/S² FOR 5 MIN. CALCULATE THE SPEED ACQUIRED BY TRAIN AND DISTANCE TRAVELLED BY TRAIN BEFORE STOPPING​​

Answer 1

Answer:

Initial velocity, u=0m/s

Final velocity, yv=?

Acceleration, a=0.2m/s

2

Time, t=5min=5×60=300sec

Using first equation of motion to obtain the final speed:

v=u+at

v=0+0.2×300=60m/s

And the distance travelled is

s=ut+

2

1

αt

2

s=0×300+

2

1

×0.2×300×300

s=0+9000=9000m=9km

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Answer 2

Answer:

u = 0 m/s

a = 0.2 m/s²

t = 5 min = 5 × 60 sec = 300 s

v = u ÷ at

v = 0 + 0.2 × 300

v = 60 m/s

s = ut + 1/2 at²

s= 0 + 1/2 0.2 × 300 × 300

s = 9000 m