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I calculated (a)
I don't know how to calculate (b)
Thank you so so so so much! That was so so helpful!after integrating both of them you will get this
Can you also pls tell how to calculate 7(c).after integrating both of them you will get this
Thank you very much!For $0 \le t \lt 1$ $s= 1/2 t^2 + 2/3 t^3$ putting t=1 we get 7/6
For $1 \le t \le 5 $ $v= 30/4 t + 5/4 t^{-2} - 23/4 $ integrate and putting limits we get $x-7/6 = (15/4) (5^2 - 1^2) + (5/4)(1- 1/5)+ (23/4)(1-5)$ I'm getting x =63 + $7 \over 6$ So total distance = 63 + $7 \over 3$ = 65.34
Even I am also not getting the answer may be I made a calculation mistake or the answer is wrong. Afaik this will be done here.