At a distance L=400m away from the signal light, brakes are applied to a locomotive moving with a velocity, u=54km/h. determine the position of rest of the locomotive relative to the signal light after 1 min of the application of the brakes if its acceleration a= -0.3 m/s2 ??
Answers
Answered by
0
Answer:
Correct option is
C
17.69ms
−1
Total distance covered(s) = Distance during acceleration(s
1
) + distance during uniform motion(s
2
) + distance during retardation(s
3
)
For acceleration,
v=u+a∗t
v
max
=2×10=20 m/s
s=ut+(1/2)at
2
s
1
=(1/2)×2×10
2
=100 m
For uniform motion,
s=vt
s
2
=20×200
=4000 m
During retardation,
v=u+at
0=20+50t
=−0.4 m/s
2
s
3
=ut+(1/2)at
2
=20×50+(1/2)×(−0.4)×50
2
=500 m
Total distance travelled, s=s1+s2+s3=100+4000+500=4600 m
Total time taken, t=t
1
+t
2
+t
3
=260 s
Average velocity, v
avg
=
Total Time
Total Distance
=4600/260=17.69 m/s
Answered by
3
Answer:
u = 54 x 5/18 = 15 m /s a = - 0.3 m/s2 ∴ v = u + at 0 = 15 – 0.3 t0 t0 = 15 / 0.3 = 50 sec After 50 second, locomotive comes in rest permanently. ∴ the distance of the locomotive from traffic light = 400 – 375 = distance-400m-from-the-traffic-light-brakes-are-applied-locomotive-moving-velocity-54-km
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