Question

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 the distance that the particle traveled on this time interval?

Answer 1

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