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Suppose that a particle moves along a straight line with velocity $v(t) = 7 - 4t$, where $0 \le t \le 0.5$ (in meters per second). Find the formula for the displacement of the particle and the total distance it has traveled at time $t = 0.5$ seconds.

          Suppose that a particle moves along a straight line with velocity $v(t) = 7 - 4t$, where $0 \le t \le 0.5$ (in meters per second). Find the formula for the displacement of the particle and the total distance it has traveled at time $t = 0.5$ seconds.
        
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Suppose that a particle moves along a straight line with velocity v(t) = 7 - 4t, where 0 ≤ t ≤ 0.5 (in meters per second). Find the formula for the displacement of the particle and the total distance it has traveled at time t = 0.5 seconds.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Suppose that a particle moves along a straight line with velocitv vt)= 7 -4t,where 0< t<0.5 (in meters per second). Find the formula for the displacement of the particle and the total distance it has traveled at time t0.5 seconds Displacement at time t is: Total distance traveled: meters in 0.5 seconds
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Transcript

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00:01 All right, let's say we have some function v of t equal to 10 minus 4t, which describes the velocity of a particle on a line on the time interval 0 to 0 .5.
00:12 And let's say v is in meters per second, so t the time is in seconds, and the distance is in meters.
00:19 So we want to find two things.
00:21 One is the displacement as a function of time, which we'll call delta x of t.
00:26 And the other thing we want to find is the total distance on the time interval.
00:31 Okay, so first let's start with one, right? so v is equal to dx d t, and we say delta x of t is x of t minus x of zero, right? so using our knowledge of calculus, we know that x of t minus x of zero is the integral from zero to t of dt of t.
00:48 Then we just plug in, right, 10 minus 4t.
00:51 What's the integral of 10? 10t? what's the integral of minus 4t? what's minus 2t squared? and we do it from 0 to t so that's just 10 t minus 2 t squared and we're done with that part so that's the displacement function time now they want the total distance from 0 to 0 .5 right so now we integrate from 0 to 0 .5 d t absolute value of vt because it's distance distance you always do absolute value however we're in a we're in a bit of luck here because interval is from 0 to 0 .5.
01:35 So 10 minus 4 is always positive on that interval...
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