Question

A car is driving on the Indianapolis 500 racetrack (a satellite photo is shown below). The Indy 500 racetrack is characterized by 2 straight segments of 5/8 of a mile, 2 straight segments of 1/8 of a mile, 4 curved segments each of 1/4 mile in length with constant radius of curvature. The total length of the racetrack is 2.5 miles

Answer 1

a)

Equilibrium of the car in X direction:

$\small \sum F_x = 0\Rightarrow F_c - Nsin\alpha-fcos\alpha = 0$

Normal Reaction:

N = mg

Friction force:

$\small f = \mu_s N = \mu_s N$

Centripetal force:

$\small F_c = \frac{mv^2}{R}$

mu2 * sz, na

Equilibrium of the car in Y direction:

$\small \sum F_y = 0\Rightarrow Ncos\alpha-fsin\alpha -mg= 0$

$\small \Rightarrow N*cos\alpha-\mu_s N*sin\alpha-mg = 0$

ng

We already have:

mu2 * sz, na

$\small \frac{mv^2}{R} = N*(sin\alpha+\mu_s cos\alpha )$

mg( sina+Hscosa

with given values:

R = 1/4 mile = 402.34 m

9.81 2 * (sin9 + ) 0,7 k cos9 402.34 cos9 -0.7sin9

$\small \Rightarrow v = 61.729 m/s$

which is equivalent to :

138.1mph

b)

Time taken for 4 curved segments is:

dist time: t

$\small time:t = \frac{4*402.34}{61.729}$

time:t- 26.071sec

Time for straight segments:

First acceleration and then deceleration.

Both initial and final velocities are same and equal to $\small \Rightarrow v = 61.729 m/s$

acceleration:

$\small v = u+at_1$

$\small v_a = 61.729+19.6*t_1$

deceleration:

$\small v = u-at_2$

$\small 61.729 = u_d - 19.6*t_2$

But final velocity for acceleration is initial velocity for deceleration.

$\small v_a = u_d$

total distance is maximum if both acceleration and deceleration are for equal time.

for 5/8 mile = 1005.84m

$\small 1005.84 = 61.729*t_1+\frac{1}{2}19.6*t_1^2+ v_a*t_2-\frac{1}{2}19.6*t_1^2$

$\small 1005.84 = 61.729*t_1+\frac{1}{2}19.6*t_1^2+ (61.729+19.6t_1)*t_2-\frac{1}{2}19.6*t_2^2$

$\small t_1 = t_2 = t$

$\small 1005.84 = 61.729*t+\frac{1}{2}19.6*t^2+ (61.729+19.6t)*t-\frac{1}{2}19.6*t^2$

$\small 1005.84 = 123.458*t+19.6t^2$

solving it, we get:

$\small t = 4.676s$

total time is:(twice - for acceleration and deceleration)

$\small t = 2*4.676s = 9.352 sec$ this is for 1 5/8 mile track

for 1/8 mile = 201.17m

$\small 201.17 = 61.729*t_1+\frac{1}{2}19.6*t_1^2+ v_a*t_2-\frac{1}{2}19.6*t_1^2$

19.6 * tỉ + (61.7294 19.64) * t2--19.6 * t 201. 17 = 61.729 * t1 +

$\small t_1 = t_2 = t$

$\small 201.17 = 61.729*t+\frac{1}{2}19.6*t^2+ (61.729+19.6t)*t-\frac{1}{2}19.6*t^2$

$\small 201.17 = 123.458*t+19.6t^2$

solving it, we get:

$\small t = 1.343s$

total time is:(twice - for acceleration and deceleration)

$\small t = 2*1.343s = 2.686 sec$ this is for 1 5/8 mile track

Maximum speed:

$\small V = 61.729+19.6*4.676$

$\small V = 153.3786 m/s$

which is equivalent to 343.098 mph

Total time is:

$\small T = 2*9.352+2*2.686+26.071$

$\small T = 50.147s$ We don't break the record