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Verizon [17]
3 years ago
14

The velocity of a particle moving along a line is given by v(t) = e ^ t * sin(e ^ t) the interval 0<= t<= pi 2 . What is t

he distance that the particle traveled on this time interval?
Mathematics
1 answer:
Butoxors [25]3 years ago
4 0

Answer:

0.44237 units

Step-by-step explanation:

Given

v(t) = e^t*sin(e^t)

0 \le t \le \frac{\pi}{2}

Required

The distance traveled in this interval

We have:

v(t) = e^t*sin(e^t)

The distance is calculated as:

d(t) = \int\limits^a_b {v(t)} \, dt

So, we have:

d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(e^t)} \, dt

Let:

u = e^t

Differentiate

\frac{du}{dt} = e^t

So:

dt = e^{-t} du

So, we have:

d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(e^t)} \, dt

d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(u)} \, e^{-t} du

Rewrite as:

d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t* e^{-t}*sin(u)} \, du

d(t) = \int\limits^{\frac{\pi}{2}}_0 {sin(u)} \, du

Integrate:

d(t) = -\cos(u)|\limits^{\frac{\pi}{2}}_0

Substitute u = e^t

d(t) = -\cos(e^t)|\limits^{\frac{\pi}{2}}_0

Split

d(t) = -\cos(e^\frac{\pi}{2}) - [-\cos(e^0)]

d(t) = -\cos(e^\frac{\pi}{2}) +\cos(e^0)

d(t) = -0.09793 + 0.5403

d(t) = 0.44237

The distance traveled in this interval is: 0.44237 units

dt =

d(t) = \int\limits^2_0 {e^t*sin(e^t)} \, dt

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