The velocity function is v(t) = t^2 - 6 t + 8 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-1,6].

Answer 1

v=ds/dt

ds/dt=t^2-6t+8

ds=(t^2-6t+8)*dt

s=integral(t^2-6t+8)*dt

s=(t^3-9t^2+24t)/3+C

s=(t/3)*(t^2-9t+24)+C

s=(t/3)*[(t-9)*t+24)]+C

s=Definite integral betwen(-1,6)

s=(6/3)*[(6-9)*6+24]-(-1/3)*[(-1-9)*(-1)+24]

s=(6/3)*[(-3)*6+24]+(1/3)*[-10*(-1)+24]

s=2*(-18+24)+(1/3)*(10+24)

s=2*6+(1/3)*34
s=12+34/3

s=(36/3)+(34/3)

s=70/3