A particle travels along a straight line with a velocity. What is the total distance?

solve for v(t)=0 over the interval [0,2]

sin(pi)<0, try t=pi/2,

3e^(-pi/4)sin(pi)= 0, so the interval 0 to pi/2 the particle travels forward [0,pi/2), then it goes negative for the rest of the interval (pi/2,2) and travels backward.

total distance = |integral 0 to pi/2 v(t) dt| + |integral pi/2 to 2 v(t) dt|

What you've calculated is displacement: ∫₀² v(t) dt = 1.8495

http://www.wolframalpha.com/input/?i=%E2%88%AB+3e^...

Displacement is difference between end position and start position. Displacement can be negative (if end position is to the left of start position) or positive (if end position is to the right of start position)

Distance is always positive, and what the particle actually travelled. (ex: if you drive 2 miles to the store and drive 2 miles home, displacement = 0, but distance travelled = 4 miles)

Notice the area under the graph in link above:

The blue region is a positive area and indicates displacement of particle travelling in positive direction (i.e. to the right), while pink region is a negative area and indicates displacement of particle travelling in negative direction (i.e. to the left)

So integrating from 0 to 2 will add positive area and negative area, or it will subtract absolute value of area below x-axis from value of area above x-axis.

But to find distance travelled, you want to add absolute area of both regions.

Example:

Particle travels to the right 5 units (area = 5)

Particle travels to the left 3 units (area = −3)

Total displacement = 5 − 3 = 2

Total distance travelled = 5 + 3 = 8

What you need to calculate is: ∫₀² | v(t) | dt

Now if you use a graphing calculator this is simple.

Just add absolute value around the function: ∫₀² | v(t) | dt = 2.26138

http://www.wolframalpha.com/input/?i=%E2%88%AB+|+3...

Notice how both areas are now both above the x-axis?

But when you do this by hand, you need to find the point where velocity > 0 and where velocity is < 0. So first we find where v = 0

3 e^(−t/2) sin(2t) = 0

e^(−t/2) = 0 -----> no solution

sin(2t) = 0 -----> 2t = 0 or π -----> t = 0 or π/2

On interval (0, π/2) ----> v(t) > 0 ----> ∫ | v(t) | dt = ∫ v(t) dt

On interval (π/2, 2) ----> v(t) < 0 ----> ∫ | v(t) | dt = − ∫ v(t) dt

Distance travelled = ∫[0 to π/2] v(t) dt − ∫[π/2 to 2] v(t) dt

∫[0 to π/2] v(t) dt = ∫[0 to π/2] (3 e^(−t/2) sin(2t)) dt

. . . . . . . . . . . . . = ∫[0 to π/2] (3 e^(−t/2) sin(2t)) dt

. . . . . . . . . . . . . = −6/17 e^(−t/2) (sin(2t) + 4cos(2t)) |[0 to π/2]

. . . . . . . . . . . . . = −6/17 (e^(−π/4) (sin(π) + 4cos(π)) − e^(0) (sin(0) + 4cos(0)))

. . . . . . . . . . . . . = −6/17 (e^(−π/4) (0 − 4) − (0 + 4))

. . . . . . . . . . . . . = −6/17 (−4e^(−π/4) − 4)

. . . . . . . . . . . . . = 24/17 (e^(−π/4) + 1)

. . . . . . . . . . . . . = 2.05544

∫[π/2 to 2] v(t) dt = ∫[π/2 to 2] (3 e^(−t/2) sin(2t)) dt

. . . . . . . . . . . . . = ∫[π/2 to 2] (3 e^(−t/2) sin(2t)) dt

. . . . . . . . . . . . . = −6/17 e^(−t/2) (sin(2t) + 4cos(2t)) |[π/2 to 2]

. . . . . . . . . . . . . = −6/17 (e^(−1) (sin(4) + 4cos(4)) − e^(−π/4) (sin(π) + 4cos(π)))

. . . . . . . . . . . . . = −6/17 ( (sin(4)+4cos(4))/e − e^(−π/4) (0 − 4))

. . . . . . . . . . . . . = −6/17 ( (sin(4)+4cos(4))/e + 4e^(−π/4))

. . . . . . . . . . . . . = −0.205938

So when you add the two areas, you get displacement:

2.05544 + (−0.205938) = 2.05544 − 0.205938 = 1.849502

This is what you found

But when you add absolute values of area you get actual distance travelled:

∫[0 to π/2] v(t) dt − ∫[π/2 to 2] v(t) dt

= 2.05544 − (−0.205938)

= 2.05544 + 0.205938

= 2.261378

Answer: (d)

Mαthmφm

(b) 1.850