A particle moves in a straight line with velocity function v(t) = 2√t+3 cms¯¹, t > 0. a Find the average acceleration from t = 1 to t = 4 seconds.
Answers 2
V=∣t−4∣
V=t−4 for t>4
V=−(t−4) for t<4
For t∈[0,4]
v=−t+4
a=−1m/s2
Since acceleration is uniform at t=0,v=4m/s
x=21×1×16
x=8m
For t>0, v=t−4
Acceleration =1m/s
s=21×4×4=8m
Total distance =16m
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Author:
chiczru3
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8
Correct option is B)
V=∣t−4∣
V=t−4 for t>4
V=−(t−4) for t<4
For t∈[0,4]
v=−t+4
a=−1m/s
2
Since acceleration is uniform at t=0,v=4m/s
x=
2
1
×1×16
x=8m
For t>0, v=t−4
Acceleration =1m/s
s=
2
1
×4×4=8m
Total distance =16m
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Author:
baby boo
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Rate an answer:
6
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