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Sagot :
The car reaches a maximum speed of 36 m/s. It accelerates for 20 seconds, reaching this speed, and then decelerates for the next 20 seconds back to rest.
The motion curves:
Acceleration: Constant at 1.8m/s21.8m/s 2 for 20seconds, then −1.8 m/s2−1.8m/s 2 for the next 20 seconds.
Velocity: Increases linearly to 36 m/s in 20 seconds, then decreases linearly back to 0 m/s.
Position: A smooth quadratic curve, representing the total distance of 540 meters.
Answer:
- Maximum speed: 18 m/s
- See the attachment for the motion curves
Step-by-step explanation:
You want the maximum speed and the motion curves for a car that travels 540 m from A to B in exactly 40 s, accelerating and decelerating at 1.8 m/s².
Velocity curve
The velocity curve will show velocity increasing at 1.8 m/s² to some maximum speed, holding at that speed for a period of time, then decelerating at -1.8 m/s² to reach zero speed at a distance of 540 m. The distance is the area under this trapezoidal shaped velocity curve.
The time to reach maximum velocity, or to decelerate from maximum velocity will be ...
[tex]t_1=\dfrac{v_{max}}{a}=\dfrac{v_{max}}{1.8}[/tex]
Distance
The total distance traveled during the acceleration and deceleration phases of the trip will be ...
[tex]d_1+d_2=\dfrac{v_{max}}{2}t_1+\dfrac{v_{max}}{2}t_2=v_{max}\cdot t_1\qquad\text{for $t_1=t_2$}\\\\\\d_1+d_2=v_{max}\cdot\dfrac{v_{max}}{1.8}=\dfrac{v_{max}^2}{1.8}[/tex]
The distance traveled during the period of constant velocity will be ...
[tex]d_3=v_{max}\cdot(40-2t_1)=v_{max}\cdot\left(40-2\cdot\dfrac{v_{max}}{1.8}\right)\\\\\\d_3=40\,v_{max}-\dfrac{2v_{max}^2}{1.8}[/tex]
We require the total distance to be 540 meters, so ...
[tex]d_1+d_2+d+3=540=\dfrac{v_{max}^2}{1.8}+40\,v_{max}-\dfrac{2v_{max}^2}{1.8}\\\\\\540=40\,v_{max}-\dfrac{v_{max}^2}{1.8}\\\\v_{max}^2-72\,v_{max}+972=0\qquad\text{put the quadratic in standard form}\\\\(v_{max}-18)(v_{max}-54)=0[/tex]
The relevant solution to this equation is ...
[tex]v_{max}=18\qquad\text{meters per second}[/tex]
The maximum velocity is 18 m/s.
Motion curves
As shown in the attachment, this value of maximum velocity means the velocity (and speed) curves show linearly increasing velocity at 1.8 m/s² for 10 seconds, to a maximum velocity of 18 m/s. That speed is maintained for 20 seconds before the car decelerates at the constant rate of -1.8 m/s² until it comes to rest after 40 seconds at a distance of 540 meters from its starting position.
The acceleration is a step function, having a value of +1.8 m/s² for 10 seconds, 0 for the next 20 seconds, and -1.8 m/s² for the final 10 seconds.
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