MHB General motion in a straight line

[Shah 72]

I calculated (a)
I don't know how to calculate (b)

 

[DaalChawal]

after integrating both of them you will get this

 

[Shah 72]

after integrating both of them you will get this

Thank you so so so so much! That was so so helpful!

 

[Shah 72]

after integrating both of them you will get this

Can you also pls tell how to calculate 7(c).
I calculate s1 for interval between t=0 and t=1 and s2 for the interval between t=1 and t=5 by integrating their respective velocities. I did that and I get the ans 30m but the textbook ans is 69.2m. Pls help

 

[DaalChawal]

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.

 

[Shah 72]

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.

Thank you very much!