Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7
How to solve for the value of s2 - s1
We have
= [tex]\frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt[/tex]
[tex]= \int\limits^a_b {(5-3t)} \, dt[/tex]
[tex]5t - \frac{3t^2}{2} +c[/tex]
v2 = 5x2 - 3x2 + c
= 10-6+c
= 4+c
[tex]s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c[/tex]
S2 - S1
[tex]=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)[/tex]
= 6 + 6+c - 2+3+c
12+c-5+c = 0
7 = c
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