Question

Total: 20 pts) A car is braking hard. There are two significant resistive forces acting on it, a quadratic (cv2) air drag, and a constant (umg) frictional force. If you are interested in finding v(x) (rather than v(t)), there is a commonly used method known as the "v dv/dx rule”, which uses the chain rule to rewrite ü = = di = v . a) [12 pts) Write down the equation of motion for Ü = (v) and use the "v dv/dx” rule to solve the equation of motion directly for v(x), and show that the distance the car needs for a full stop is: A2 A2 + 2 max = 2... In 2ug Here, u is the friction coefficient. What is the constant A in this case (in terms of given parameters in the differential equation)? b) 8 pts The SSC Ultimate Aero TT, one of the world's fastest production cars (at a cool $600,000+), has a maximum speed of 412 Km per hour. The engine provides maximum forward force of 11260 N on the 1200 kg car. For this car, c =0.86 kg/m. On a race track, the car exits a turn with a speed of vo = 300 km/hr. As soon as the driver enters the straight track after the turn, she realizes there is another car blocking the track 2 km away. The driver slams on the brakes. Assuming a friction coefficient p = 0.7, use your above result to compute the distance required for the Aero TT to stop. Compare this distance to the corresponding distance in the absence of air drag (i.e. just kinetic road friction). Briefly, discuss.

Answer 1

Answer:

Given that:

A car is braking hard.There are two significant resistive force acting on it, a quadratic (cv²) air drag , and a constant ($\mu mg$)frictional force.

a) First we write down the equation of motion for the car

m = -cu2 – umg (1) dt

The minus sign means that the forces are resistive. Dividing by m and using the chain rule as mentioned in the question , we get

do a 04+ 402 = -(2) drm

Multiplying equation (2)  by 2, we get

20 +2-1+ = -2ug → dcm v e-2) = -2uge — (3) m m

Integrating equation (3)om both sides

- 2c - 22, 2c mg 2c ve- = - umga er +D= Dex= 12 + m

D is the integration constant that we can determine by the condition that before the car hits breaks its velocity is v₀ and x=0 . This gives us

umg D = 3 +

To find X_max we set v=0 in equation (4).

buri ш el mar 2c Dº phim

Defining A² = pang equation (5)becomes

$x_{max}=\frac{m}{2c}In\left ( \frac{A^{2}+v_{0}^{2}}{A^{2}} \right )=\frac{A^{2}}{2\mu g}In \left ( \frac{A^{2}+v_{0}^{2}}{A^{2}} \right )(6)$

b)The initial velocity of the car is v₀ =300km/hr =83.33ms^-1 , the constant is

22 0.7 * 1200kg 9.8ms-1 0.86kgm-1 -=9572m2-2

Hence, using equation (6), x_max =380m.If there was no air drag the equation of motion will be

du V dr =-ug gx = D -

Again using x = 0 when v = v₀ ,we get

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putting in the given values we get x_max =506m. As expected the distance required to stop the car without the air drag is large.