Question

The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into the indicated number of subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles. v = 1/(2t + 4) (m/s) for for 0 ≤ t ≤ 88; n = 22

Answer 1

Answer:

The displacement of the object on this intervals is 1.33 m.

Step-by-step explanation:

Given that,

The function of velocity is

$v=\dfrac{1}{2t+4}\ m/s$

For 0 ≤ t ≤8 , n = 2

We need to calculate the intervals

Using formula for intervals

For, n = 1

$\Delta x=\dfrac{t_{f}-t_{i}}{n}$

$\Delta x=\dfrac{8-0}{2}$

$\Delta x=4$

So, The intervals are (0,4), (4,8)

We need to calculate the velocity

Using given function

$v=\dfrac{1}{2t+4}$

For first interval (0,4),

Put the value into the formula

$v_{0}=\dfrac{1}{2\times0+4}$

$v_{0}=\dfrac{1}{4}$

For first interval (4,8),

Put the value into the formula

$v_{4}=\dfrac{1}{2\times4+4}$

$v_{4}=\dfrac{1}{12}$

We need to calculate the total displacement

Using formula of displacement

$D=(v_{0}+v_{4})\times(\Delta x)$

Put the value into the formula

$D=(\dfrac{1}{4}+\dfrac{1}{12})\times4$

$D=1.33\ m$

Hence, The displacement of the object on this intervals is 1.33 m.

Answer 2

The displacement of the object whose velocity function is given is 1.33 m

The given parameters are:

$\mathbf{v = \frac{1}{2t + 4},\ 0 \le t \le 8; n =2}$

The end point of intervals is calculated as:

$\mathbf{\triangle t = \frac{b - a}{n}}$

So, we have:

$\mathbf{\triangle t= \frac{8 - 0}{2}}$

$\mathbf{\triangle t = \frac{8}{2}}$

$\mathbf{\triangle t= 4}$

So, the intervals are (0,4) and (4,8)

Calculate the velocity at the beginning of each interval

$\mathbf{v_0 = \frac{1}{2(0) + 4} = \frac 14}$

$\mathbf{v_4 = \frac{1}{2(4) + 4} = \frac 1{12}}$

Calculate the displacement (S) using:

$\mathbf{S = (v_0 + v_4) \times \triangle t}$

So, we have:

$\mathbf{S = (1/4 + 1/12) \times 4}$

Expand

$\mathbf{S = 1 + 1/3}$

Add

$\mathbf{S = 1 \frac 13}$

Express as decimals to 2 decimal places

$\mathbf{S = 1.33}$

Hence, the displacement is 1.33 m

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