Mechanics paper/Math (please helpp!!!)
1 answer:
Answer:
(i) 15 m, 6 m/s
(ii) 90 m
Step-by-step explanation:
(i) For some acceleration (a) from rest, the distance covered (d) in time t is ...
d = (1/2)at^2
The distance covered by Ben in the 5 seconds he is accelerating is ...
d = (1/2)(1.2 m/s²)(5 s)² = 15 m
Of course, Ben's speed at that point is ...
s = (1.2 m/s²)(5 s) = 6 m/s
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(ii) When Ben has been walking 5 s, Alan has been walking 10 s, so Alan has covered (10 s)(4 m/s) = 40 m. Their distance difference of 40 -15 = 25 m is being made up at the rate of their speed differences: (6 m/s) -(4 m/s) = 2 m/s.
It will take (25 m)/(2 m/s) = 12.5 s additional time for Ben to catch Alan. In the 22.5 s that Alan has been walking before they meet, he will have walked ...
(22.5 s)(4 m/s) = 90 m . . . the distance OP
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Answer:
D)Yes, because the difference in the means in the actual experiment was more than two standard deviations from 0.
Step-by-step explanation:
We will test the hypothesis on the difference between means.
We have a sample 1 with mean M1=18.2 (drug group) and a sample 2 with mean M2=15.9 (no-drug group).
Then, the difference between means is:
If the standard deviation of the differences of the sample means of the two groups was 1.1 days, the t-statistic can be calculated as:
The critical value for a two tailed test with confidence of 95% (level of significance of 0.05) is t=z=1.96, assuming a large sample.
This is approximately 2 standards deviation (z=2).
The test statistict=2.09 is bigger than the critical value and lies in the rejection region, so the effect is significant. The null hypothesis would be rejected: the difference between means is significant.