 For the second problem we'll analyse a car accelerating from a stop position at a traffic light. Again the first thing we do is draw a free body diagram. Here we have gravity acting downwards and the normal force acting upwards. When you press on the accelerator this causes the wheels to rotate and the tyre pushes backwards on the road. Because the tyre is not slipping on the road the part of the tyre that is touching the road is actually stationary for the short time interval that it's in contact with the road, otherwise it would be freely rotating and you would lose traction. This means that it's static friction rather than kinetic friction that is relevant here. As the tyre pushes back on the road the static friction force pushes the car forward. This might seem counterintuitive but if you imagine a car on ice the wheels would rotate and spin freely against the ice and the car would not move forward. Let's calculate the maximum acceleration that's possible without spinning the tyres. The maximum acceleration will occur when the frictional force is maximum. Now the largest the frictional force can be is mu s n. This problem is more complicated than the last one because we have forces parallel to the road and forces perpendicular to the road. What we need to do is to write a separate f equals ma equations in both the x and y directions. In the x direction the only force acting is friction and so f net in the x direction is equal to mu s n. In the y direction we have the normal force pointing in the positive y direction and gravity pointing in the negative y direction. If we add these up we get the net force in the y direction is equal to n minus mg and because the car is not accelerating upwards or downwards the acceleration in the y direction is zero and so f net in the y direction is equal to zero. This allows us to solve for n which we can then substitute back into our force equation in the x direction. We then apply a Newton's second law in the x direction f net equals ma and find that the maximum possible acceleration of the car without spinning the tyres is equal to mu s g. You'll notice that we've solved all of this algebraically without substituting any numbers in. It can be really tempting when you're given numbers in the question to substitute them into the equation right at the start. The problem with this is that it's much harder to check that your answer makes sense. The easiest check to do is making sure that the dimensions on both sides of your final equation match. Here we have acceleration which has units of meters per second squared equals to mu s the coefficient of friction which has no dimensions and g which also has units of meters per second squared so we're all good to now go ahead and start substituting numbers into the answer. For a standard car on a road mu s is around 0.9 which gives us an acceleration of 8.8 meters per second squared. This means that every second we can increase the speed of the car by 8.8 meters per second. Let's briefly compare this to the acceleration if you do spin the tyres also known as burning rubber. If the tyres are moving relative to the road surface then it's the kinetic not static friction force that's pushing the car forward. Basically all the forces are the same so we can just replace the mu s with mu k. For a standard car mu k is around 0.7 so this gives us a smaller acceleration 6.9 meters per second squared so it's definitely less effective to accelerate when you spin the tyres.