# The velocity is v(t)=−t2+6t−8 for a particle moving along a line. Find the the distance traveled during the time interval [-3,6].?

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• I'll show you how to do what @andrew suggested.

Find where the velocity changes sign:

v = 0 = -t² + 6t - 8

This is quadratic with roots at t = 2 and t = 4

At t = 3, v = -3² + 6*3 - 8 = 1

which is positive

and so the velocity must be negative for t < 2 and t > 4.

s(t) = ∫ v(t) dt = -t³/3 + 3t² - 8t + C

Evaluated for -3 < t < 2, s = -66.7

and for 2 < t < 4, s = 1.33

and for 4 < t < 6, s = -6.67

and so the total DISTANCE is

d = 66.7 + 1.33 + 6.67 = 74.7  ◄

(meters, presumably)

Note that if you do a single integration from -3 to 6 you get -72 (meters, presumably). That's the DISPLACEMENT during the time interval.

• NCS
Lv 7
1 month agoReport

Hopefully you don't think that I gave you a thumb down. I did not.

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• Anonymous
1 month ago

72 m....................

• Andrew Smith
Lv 7
1 month agoReport

This is the absolute value of the DISPLACEMENT, but it isn't the distance moved.

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• Anonymous
1 month ago

total distance is ≅ 183.5 m

• Andrew Smith
Lv 7
1 month agoReport

No it isn't.  You might like to give a reason why you chose this value.

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• Because they asked for DISTANCE ( not displacement) you must determine the times ( if any) at which it changes direction.  ( 2 and 4 seconds) now integrate. s(t) = -t^3/3 + 3t^2 -8t (+c)

Find how far it moves in each of the three intervals  and add the absolute values of each of these motions.

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