19. A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it covers
period? Assume that the tracks are straight?
Answers
Answered by
0
Given:
Initial velocity = u = 10 km/h = 2.7 m/s
Final velocity = v = 40 km/h = 11.1 m/s
Time = t = 2 minutes = 120s
To find:
Distance covered = s = ?
Explanation:
Accelaration = a = v-u/t = 11.1 - 2.7/ 120 = 0.07 m/s^2
Using kinematical eq 2,
s = ut + 1/2 at^2
s = 2.7*120 + 1/2 * 0.07 * 120^2
s = 324 + 1/2 * 0.07 * 14400
s = 324 + 1/2 * 7 * 144
s = 324 + 7 * 72
s = 324 + 504
Therefore, distance covered is 828m or 0.828km.
Answered by
0
Answer:
u=10 kmhr
v=40kmhr
t=2min=2×60=120second
a=v-u/t
=40-10/120
=30/120
=1/4 ms^2
s=ut+ 1/2at^2
=10×120+1/2×1/4×120×120
=1200+60×30
=1200+1800
=3000m
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