question 1: a vehicle’s velocity in km/hr is given as $v = c_0 + c_1t - c_2t^2$. where $t$ is in seconds. if…

question 1: a vehicle’s velocity in km/hr is given as $v = c_0 + c_1t - c_2t^2$. where $t$ is in seconds. if the variables $c_0 = 0.85 \\frac{km}{hr}, c_1 = 20.0 \\frac{km}{hr}/s, c_2 = 0.90\\frac{km}{hr}/s^2$, what is the maximum velocity of the car in $\\frac{m}{s}$? also, find the displacement from its initial position when it reaches this maximum velocity. assume the initial position is at the origin (ie., $s_0 = 0 m$) when $t = 0$. one m/s is 3.6 km/hr.

question 1: a vehicle’s velocity in km/hr is given as $v = c_0 + c_1t - c_2t^2$. where $t$ is in seconds. if the variables $c_0 = 0.85 \\frac{km}{hr}, c_1 = 20.0 \\frac{km}{hr}/s, c_2 = 0.90\\frac{km}{hr}/s^2$, what is the maximum velocity of the car in $\\frac{m}{s}$? also, find the displacement from its initial position when it reaches this maximum velocity. assume the initial position is at the origin (ie., $s_0 = 0 m$) when $t = 0$. one m/s is 3.6 km/hr.

Answer

Explanation:

Step1: Find the time when velocity is maximum

The velocity function is $v = c_0 + c_1t - c_2t^2$. To find the maximum of this quadratic - function, we take the derivative with respect to $t$ and set it equal to zero. The derivative $v^\prime=\frac{dv}{dt}=c_1 - 2c_2t$. Setting $v^\prime = 0$, we get $t=\frac{c_1}{2c_2}$. Given $c_1 = 20.0\frac{km}{hr\cdot s}$ and $c_2 = 0.90\frac{km}{hr\cdot s^2}$, then $t=\frac{20.0}{2\times0.90}\ s=\frac{20.0}{1.8}\ s\approx11.11\ s$.

Step2: Calculate the maximum velocity

Substitute $t = \frac{c_1}{2c_2}$ into the velocity function $v = c_0 + c_1t - c_2t^2$. First, convert the constants to SI - units. $c_0 = 0.85\frac{km}{hr}=\frac{0.85\times1000}{3600}\ m/s\approx0.236\ m/s$, $c_1 = 20.0\frac{km}{hr\cdot s}=\frac{20.0\times1000}{3600}\ m/s^2\approx5.56\ m/s^2$, $c_2 = 0.90\frac{km}{hr\cdot s^2}=\frac{0.90\times1000}{3600}\ m/s^3 = 0.25\ m/s^3$. $v_{max}=c_0 + c_1\times\frac{c_1}{2c_2}-c_2\times(\frac{c_1}{2c_2})^2$. $v_{max}=0.236+5.56\times\frac{5.56}{2\times0.25}-0.25\times(\frac{5.56}{2\times0.25})^2$. $v_{max}=0.236 + 5.56\times11.12-0.25\times123.65$. $v_{max}=0.236+61.827 - 30.912$. $v_{max}=31.15\ m/s$.

Step3: Find the displacement

The velocity function in SI - units is $v(t)=0.236 + 5.56t-0.25t^2$. The displacement $s$ is the integral of the velocity function with respect to time. $s=\int_{0}^{t}(0.236 + 5.56t-0.25t^2)dt$. Using the integral rules $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $s=\left[0.236t+\frac{5.56t^2}{2}-\frac{0.25t^3}{3}\right]_0^{11.11}$. $s=0.236\times11.11+\frac{5.56\times(11.11)^2}{2}-\frac{0.25\times(11.11)^3}{3}$. $s = 2.62+343.78 - 114.59$. $s = 231.81\ m$.

Answer:

The maximum velocity of the car is approximately $31.15\ m/s$ and the displacement from its initial position when it reaches this maximum velocity is approximately $231.81\ m$.