A particle moves along a line so that its velocity at time t is v(f) = t^2 - t -

Question

A particle moves along a line so that its velocity at time t is v(f) = t^2 - t - 12 (measured in meters per second). Find the displacement of the particle during 2 lessthanorequalto t lessthanorequalto 10. Find the distance traveled during this time period. By this equation, the displacement is s(10)-s(2) = integral_2^10 v(t) dt This means that the particle moved approximately: 86.67 meters to the right. Note that v(f) = t^2 - t - 12 = (t - 4)(t + 3) and so v(t) ? 0 on the interval (2, 4] and v(t) ? 0 on (4, 10). Thus, from this equation, the distance traveled is integral_2^10 |v(t)| dt = integral_2^4 [-v(t) dt + integral_4^10 v(t) dt

Explanation / Answer

a) displacement =[2 to 10]v(t)dt

=[2 to 10](t2-t -12)dt

=[2 to 10](1/3)t2-(1/2)t2 -12t

=[(1/3)102-(1/2)102 -12*10]-[(1/3)22-(1/2)22 -12*2]

=560/3

=186.67

b)v(t)=(t2-t -12)=(t+3)(t-4), and so v(t)<0 on interval [2,4] and v(t)>0 on [4,10]

[2 to 10] |v(t)|dt

=[2 to 4] -(t2-t -12)dt +[4 to 10](t2-t -12)dt

=[2 to 4] (-t2+t +12)dt +[4 to 10](t2-t -12)dt

=[2 to 4] (-1/3)t3+(1/2)t2 +12t +[4 to 10](1/3)t3-(1/2)t2 -12t

=(34/3 )+198

=628/8

=209.33

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