Question

A vehicle moves with a velocity, v(t) = exp(0.2t) - 1, 0 ≤ t ≤ 5 s. Peter would like to calculate the displacement of the vehicle as a function of time, x(t), by integrating given velocity over the time from t = 0. Use t = 0.2 s for trapezoidal rule.

Answer 1

Answer:

$x|_0^{0.2}=1.59535$

Explanation:

Given expression of velocity:

$v(t)=10^{0.2t}-1 ;\ \ 0\leq t\leq 5\ s$

For getting displacement we need to integrate the above function with respect to t.

Given period of integration:

$t_0=0\ s \to t_f=0.2\ s$

For trapezoidal rule we break the given interval into two parts of 0.1 s each.

∴take n=2

hence, $\Delta t= 0.1$

$v(0)=0$

$v(0.1)=1.0471$

$v(0.2)=1.0965$

Now, using trapezoidal rule:

$\int_{0}^{0.2}v(t)\ dt=\Delta x[\frac{1}{2}\times v(0)+v(0.1)+\frac{1}{2}\times v(0.2)]$

$\int_{0}^{0.2}v(t)\ dt=0.1 [\frac{1}{2}\times 0+1.0471+\frac{1}{2}\times 1.0965]$

$x|_0^{0.2}=1.59535$

Note:Smaller the value of sub-interval better is the accuracy.