Question

a car initially going 56 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 8 seconds. graph the velocity for ( t = 0 ) to ( t = 8 ). how far does the car travel before stopping? distance = (include units) how far does the car travel before stopping if its initial velocity is doubled, but it brakes at the same constant rate? distance = (include units)

Explanation:

Step1: Find the acceleration

The acceleration \(a\) is given by the formula \(a=\frac{v - v_0}{t}\), where \(v = 0\) (final velocity), \(v_0=56\) ft/sec (initial velocity), and \(t = 8\) s.

$$a=\frac{0 - 56}{8}=- 7\space\text{ft/sec}^2$$

Step2: Find the velocity function

The velocity function \(v(t)\) is \(v(t)=v_0+at\). Substituting \(v_0 = 56\) and \(a=-7\), we get \(v(t)=56-7t\)

Step3: Find the distance using the area under the velocity - time graph

The velocity - time graph is a straight line with \(v(0) = 56\) and \(v(8)=0\). The area under the line (a triangle) is \(A=\frac{1}{2}\times\text{base}\times\text{height}\). Here, base \(t = 8\) and height \(v_0=56\)

$$A=\frac{1}{2}\times8\times56 = 224\space\text{ft}$$

For the second part, if \(v_0'=2\times56 = 112\) ft/sec. First, find the time to stop. Using \(v = v_0'+a't\) (where \(a'=-7\) ft/sec², same acceleration). \(0 = 112-7t\), so \(t=\frac{112}{7}=16\) s

Step4: Find the distance for the new initial velocity

The area under the new velocity - time graph (a triangle with base \(t = 16\) and height \(v_0'=112\)) is \(A'=\frac{1}{2}\times16\times112\)

$$A'=\frac{1}{2}\times16\times112=896\space\text{ft}$$

Answer:

First distance: \(224\space\text{ft}\)
Second distance: \(896\space\text{ft}\)