At the beginning 3

At the beginning of motion, a car was going at a speed of 120 km/h and kept this speed for the first 11 seconds. Then it began to accelerate further such that every second it accelerated by 6 km/h until it reached a speed of 150 km/h. Then it began to decelerate until in 15 seconds it stopped.

a) Draw a graph of the motion of the car, determine the domain of definition and the range of the function, and the equations of all parts
b) Determine at what time the car had a speed equal to 140 km/h
c) Determine at what time the car had a speed less than 50 km/h
d) Determine at what time the car had a speed greater than 90 km/h
e) Determine at what time the car had a speed greater than 130 km/h

Final Answer:

b₁ =  14.3333 s
b₂ =  17 s
c =  26 s
d =  22 s
x =  12.6667 s
y =  18 s

Step-by-step explanation:

$$\begin{gathered} v=120;0s\le t\le 11s \\ v=120+6(t-11);11\le t\le 16s \\ v=150-10(t-16);16\le t\le 31s \\ 120+6\cdot (b_1-11)=140 \\ 6b_1=86 \\ {b}_1=\frac{86}{6}=14.33333333 \\ {b}_1=\frac{43}{3}\doteq 14.333333=14.3333\text{s} \end{gathered}$$

$$\begin{gathered} 150-10\cdot (b_2-16)=140 \\ 10b_2=170 \\ {b}_2=\frac{170}{10}=17 \\ {b}_2=17=17\text{s} \end{gathered}$$

$$\begin{gathered} 150-10\cdot (c-16)=50 \\ 10c=260 \\ c=\frac{260}{10}=26 \\ c=26=26\text{s} \end{gathered}$$

$d=22=22\text{s}$

$$\begin{gathered} 130=120+6\cdot (x-11) \\ 6x=76 \\ x=\frac{76}{6}=12.66666667 \\ x=\frac{38}{3}\doteq 12.666667=12.6667\text{s} \end{gathered}$$

$$\begin{gathered} 130=150-10\cdot (y-16) \\ 10y=180 \\ y=\frac{180}{10}=18 \\ y=18=18\text{s} \end{gathered}$$

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