What are thesimple physics behind driving and driving safely? Please answer soon using kinematics and dynamics

Question

I have been working my grade 11 physics project regarding the making of a slide show on physics of safe driving. I keep reading over notes and assignments and trying to incorporate what I have learned in kinematics (ie the kinematics equations) and dynamics. But it's just so hard for me. So far I have a few good slides. How do I incorporate subjects like the force of gravity Newton's laws of motion and other subjects into my presentation? Please answer my question as simply as you can and as quick as you can because my project is due soon!

Answer 1

Really great question. Physics is best learned when put into the context of day to day living.

Here's what you need to show.

Time it takes to brake to a halt on good road conditions vs. bad road conditions.

* SUVAT (kinematic) equations: v = u + at; where u = is the car speed when braking starts and v = 0, end speed. Thus, t = u/a; we need to find a, the braking deceleration, which depends on the road conditions.

* Force(dynamic) equations: F = kN; where k is the coefficient of rolling (static) friction and N = mg cos(theta) the normal weight of the car of mass m on a road with incline theta degrees. And f = ma; the net force on the car in the direction it's traveling (horizontal force).

Friction force is the only given force in the horizontal direction. Thus, f = ma = k mg cos(theta) = F; and a = k g cos(theta). This is the upper limit deceleration the car can slow down at given k and the slope of the road. g = 32.2 ft/sec^2 if you're into English units; 9.81 m/sec^2 if in SI.

Put the dynamic result into the SUVAT derived equation to get t = u/(k g cos(theta)). Now find t for good conditions (theta = 0 and k = .8 for example) and bad (theta = 30 deg and k = .1). k = .8 is a dry road, k = .1 is a wet, slippery one. You could do a nice graph showing t vs k to show how time to stop increases linearly with decreasing road conditions k.

Then, lesson learned, point out how more time to stop safely means more space between you and the car ahead is necessary to ensure no rear ender. That is, put more room between you and the car ahead in bad weather.

Safe speed for going around a curve on good road conditions vs. bad road conditions.

*Force equations: P = mv^2/R; centripetal (centifugal) force with radial acceleration a = v^2/R; where v = tangential velocity, R is radius of the curve. F = kN = k mg cos(theta); this time theta is the embankment of the curved road. Then, to not skid outward and off the road, f = ma = P - F = 0 has to be true to stop acceleration outward along the radius of turn.

Thus, P = mv^2/R = ma = k mg cos(theta) = F; so that v^2 = k R g cos(theta) and v = sqrt(k R g cos(theta)). v is the maximum speed the car can go and still hold the road while in the turn around radius R on a road embanked at a theta degrees slope. To show the effects of k and theta (the road conditions) do the ratio (V/v)^2 = K R g cos(THETA)/k R g cos(theta) = (K/k)(cos(THETA)/cos(theta)); so that V^2 = v^2 (K/k)(cos(THETA)/cos(theta)).

Use this last equation, the ratio, to show that if K > k, better friction, V > v, you can go around the curve faster. Or if THETA > theta, steeper embankment, V > v and you can go faster without sliding outward into the trees.

THE SIMPLE ANSWER:

Slide 1. Show t = u/a = u/(k g cos(theta)) and graph t vs k, and t vs theta. Exaplain why this means leaving more distance between cars in bad road conditions.

Slide 2. Show V^2 = v^2 (K/k)(cos(THETA)/cos(theta)). Explain how K > k gives V > v so taking the curve at higher speeds in good conditions is feasible; and how THETA > theta has the same result (V > v).