A car has broken down in the middle of a road. Eva wants to help tow the car to a mechanic with her 4WD. To avoid the car from slamming into Eva’s 4WD, they attach the front of the car to the back of Eva’s 4WD by a stiff spring. The broken car weighs 2000kg. The spring has a natural unstretched length of 2m and a spring constant of k = 80000N/m. This means if the spring is pulled beyond its unstretched length by d metres, then it exerts a contracting force of 80000d Newtons. Conversely if the spring is compressed from its unstretched length by d metres, then it exerts an expanding force of 80000d Newtons. This is known as Hooke’s Law, and can be written as T = kd For this particular gravel road, the magnitude of the drag force is proportional to the weight of the car and the speed of the car. The direction of the drag force is always opposite to the direction of movement. This can be written as D = −cM v For this particular road we can take c = 4. Let x(t) be the position of the front of the broken car and let y(t) be the position of the back of Eva’s 4WD. We will assume that the position of Eva’s car is a known function of time. (1) By considering Newton’s law applied to the broken car, write down an ordinary differential equation for the motion of the broken car They are planning to travel at a constant velocity of 36km/hr, but they need to make sure that this is safe. They want to be sure that the spring is sufficiently stiff so that if Eva suddenly stops, the broken car does not crash into Eva’s 4WD. Assume that they are both travelling at a constant velocity of 36km/hr. (2) Using the equation from Question 1, find the distance between the two vehicles, y(t)−x(t), in this scenario. Now assume that Eva breaks so sharply that she effectively stops instantaneously at position y = 2m at time t = 0. (3) What is the broken cars position and velocity at time t = 0?
A car has broken down in the middle of a road. Eva wants to help tow the
car to a
The broken car weighs 2000kg. The spring has a natural unstretched length of 2m and a spring constant of k = 80000N/m. This means if the spring is pulled beyond its unstretched length by d metres, then it exerts a contracting force of 80000d Newtons. Conversely if the spring
is compressed from its unstretched length by d metres, then it exerts an expanding force of 80000d Newtons. This is known as Hooke’s Law, and can be written as
T = kd
For this particular gravel road, the magnitude of the drag force is proportional to the weight of the car and the speed of the car. The direction of the drag force is always opposite to
the direction of movement. This can be written as
D = −cM v
For this particular road we can take c = 4.
Let x(t) be the position of the front of the broken car and let y(t) be the position of the back of Eva’s 4WD. We will assume that the position of Eva’s car is a known function of time.
(1) By considering Newton’s law applied to the broken car, write down an ordinary differential equation for the motion of the broken car
They are planning to travel at a constant velocity of 36km/hr, but they need to make sure that this is safe. They want to be sure that the spring is sufficiently stiff so that if Eva suddenly stops, the broken car does not crash into Eva’s 4WD.
Assume that they are both travelling at a constant velocity of 36km/hr.
(2) Using the equation from Question 1, find the distance between the two vehicles, y(t)−x(t), in this scenario.
Now assume that Eva breaks so sharply that she effectively stops instantaneously at position y = 2m at time t = 0.
(3) What is the broken cars position and velocity at time t = 0?
(4) Is the spring stiff enough to stop the broken car from crashing into Eva’s 4WD?
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