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Pls help me with iii
Date Posted: 9 months ago
Views: 49
(iii) To find the distance at the 3rd second, we find the difference between the time at t=2 and t=3. So, integrate v to get s = 3t^3 - 5t^2. When t=2, s= 4 and when t =3, s= 36. So
36-4=32m
36-4=32m
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(I) Find the acceleration by differentiating v w.r.t x first. Then substitute t=2 in it to find p.
(ii) find the value of t by equating acceleration to 0, using the p value found in (i). Then substitute that t value into v.
(ii) find the value of t by equating acceleration to 0, using the p value found in (i). Then substitute that t value into v.
Date Posted:
9 months ago
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