A train starting from rest moves with a uniform acceleration of 0.2 ms-2 for 5 minutes. Calculate the final velocity and the distance travelled in this time.

Final velocity = 60 m/s

Distance travelled = 9000 m

Explanation:

Applying,

For Final velocity,

v = u+at Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, t = time.

From the question,

Given: u = 0 m/s(at rest), a = 0.2 m/s², t = 5 minutes = (5×60) = 300 s

Substitute these values into equation 1

v = 0+0.2(300)

v = 60 m/s.

Also for distance,

Applying,

s = ut+1/2at² Equation 2

Where s = distance travelled.

subtitute thes values above into equation 2

s = (0×300)+(0.2×300²)/2

s = 9000 m

6 cos 34, 6 sin 34 trig