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When driving my car over a bump or a quick change in gradient (from flat to uphill or downhill to flat), if I don't drive slowly and the bump/change in gradient is large enough, I'll end up with my front bumper dragging on the road/bump. But if I go slowly, there's no such issue.

I would expect that the wheels are still taking the same path regardless of the speed, so the car's body should also act the same regardless. Why does it make a difference when speed is changed? Why doesn't the car behave the same way regardless?

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Because your car has a suspension.

A car's wheels are not rigidly attached to the frame. Rather, they are attached with springs and shock absorbers that allow the wheels to move somewhat relative to the rest of the car. This is generally speaking a good thing, as it means that when the wheels go over a small bump or pothole in the road, the frame does not necessarily need to move up and down; and so you, the passenger, get a smoother ride. But if the bumps are large and the speeds are too high, this same "smoothing" effect means that the frame of the car can come into contact with the road surface.

As to why this smoothing effect occurs: imagine that you have, on a table, a large block (standing in for the car), attached by a spring (the suspension) to a smaller block (the wheel.) Suppose that the table is very smooth, so that we can ignore friction between the table and the smaller blocks. If you pull the small block quickly & suddenly away from the larger block, the spring stretches a lot and the large block won't move very much at all: it has a lot of inertia, and so it can't accelerate very quickly. This is the equivalent of going over a bump or pothole at high speed: the wheel suddenly moves up or down relative to the frame, but the frame doesn't move up or down much at all.

But if you pull the small block away from the large block slowly, then the large block will follow the small block, while the spring doesn't stretch terribly much. In this case, the low acceleration of the large mass takes place over a longer time, and so it can move more while the force is being exerted on it. This is the equivalent of going over a bump/pothole at low speed; since the wheels move up or down relatively slowly, the frame of the car will follow them. If you go over a bump at low speed, this means that the frame will follow the wheels (which follow the road surface), rather than moving in something resembling a straight line and possibly hitting the road surface.

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    $\begingroup$ Also worth pointing out that suspensions can behave differently depending on the stiffness of the springs. Cars that are low to the ground need to have a relatively stiff suspension to prevent bottoming out, since there's little room for the spring to travel. Vehicles that ride higher can have a softer suspension that absorbs more bumps without dragging the chassis on the ground. It's a smoother ride, but can also make the car less responsive as steering input is dampened by the shocks. $\endgroup$
    – Nuclear Hoagie
    Commented Nov 13, 2018 at 15:11
  • $\begingroup$ I really wish I could accept 2 answers. Your thought experiment is perfect, but I was hoping for a more physics answer, so I'm going with Džuris'. Thanks for the great answer!! $\endgroup$
    – scohe001
    Commented Nov 13, 2018 at 16:39
  • $\begingroup$ No, it's because the car has downforce. The suspension allows movement, but doesn't cause it. $\endgroup$
    – jpaugh
    Commented Nov 14, 2018 at 4:59
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    $\begingroup$ @jpaugh: downforce doesn't matter here. Cars are held on the road by all that heavy steel and aluminum they are made out of. $\endgroup$
    – whatsisname
    Commented Nov 14, 2018 at 5:41
  • $\begingroup$ Bear in mind that the springs absorb the shocks and the "shock absorbers" are dampers. $\endgroup$
    – Andrew Morton
    Commented Nov 14, 2018 at 11:30
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I think that physics.SE deserves an answer in physics terminology.

Change of direction = acceleration. Acceleration is caused by force. Force is applied through suspension and compresses it.

Larger speeds lead to larger acceleration and larger force which in turn compresses the spring more.

To be a bit more precise, we could model the upwards curve as consisting of arc segments. It would mean that at the segment with curvature radius $r$ the acceleration is $v^2/r$ and the force is $mv^2/r$. If we suppose the spring is linear, the compression will equal force over the spring constant - $\frac{mv^2}{rk}$.

Disclaimer: the spring is probably not linear, it should become increasingly high-degree-polynomial near the extremes of it's range (i.e. it should try to stop itself from compressing too much). The $m$ is not the mass of the car but something like the portion that rests on the axis in question. Which depends on multiple factors including the road curvature and the absolute acceleration of car. And the $r$ is not the curvature radius of the road but that of the car's body. Suspension is intended exactly to make it less than the curvature of the road - smoothen the bump.

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    $\begingroup$ Car suspension is designed to be pretty close to linear across the entire motion ratio, so it's a valid assumption to make - also +1 for being the only answer to mention the velocity squared part, as that's the key issue I see - the question's assumption about the car's wheels and body taking the same path is false $\endgroup$
    – user2813274
    Commented Nov 12, 2018 at 23:19
  • $\begingroup$ Awesome explanation. The v^2 relationship specifically explains why there's such a big difference in just a few mph. $\endgroup$
    – scohe001
    Commented Nov 13, 2018 at 16:37
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    $\begingroup$ @jpaugh: unless you are driving a purpose-built racing car at ludicrous speed, the aerodynamic downforce is not significant. Gravity is sufficient in analyzing this situation. $\endgroup$
    – whatsisname
    Commented Nov 14, 2018 at 5:38
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    $\begingroup$ This is not a good answer. It is factually correct in its statements but it does not present a complete chain of reasoning between a large speed and scraping the front into the speed bump. $\endgroup$
    – MichaelK
    Commented Nov 14, 2018 at 16:20
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    $\begingroup$ @jpaugh: I can't understand the level of confusion you're describing here. The system can be understood even if you think of a car as having just one wheel. The mass of the car resists changes in velocity. If the ground underneath moves fast enough, i.e. a big enough bump relative to the cars speed, the springs won't deliver enough force to move the car's mass and the vehicle will bottom out. Throw a few thousand pounds of bricks in a car and you can bottom it out hitting bumps at walking speeds. $\endgroup$
    – whatsisname
    Commented Nov 14, 2018 at 17:33
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In order to avoid the issue, your car needs to be lifted over the bump. In the case of fast motion, this has to occur in a shorter time span, hence with a higher acceleration, hence with greater force - and this force has to be transferred from the ground via your wheels to the car body. However, the positional relation between your tires and the car body is governed by springs and shock absorbers, which are compressed under strong forces. Additionally, your tire bottom may be compressed for the same reason. It follows that the car body moves closer to the bump than it would in slow motion.

You could avoid this phenomenon by getting rid of all springs and dampers (and also the air-filled tires) and replacing them with rigid connections. In that case, the car body would be accelerated quite suddenly with great force at the bump - probably resulting in structural failure! There's a reason why these elastic elements are built into your car after all ...

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Shocks have time factor. They will absorb more if the change is rapid. This is by design as your body and car do not want a rapid impact.

I believe springs also have a time factor.

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Hitting the bump at a higher rate of speed causes more force on the wheel and thus more force on the springs/shocks which causes the them to compress more as the wheel travels up the front side of the bump. If the bumper is far enough ahead of the wheel, it might clear the top of the bump but when the wheel goes over the top and starts going back down the backside, the springs can't push the body and bumper back up fast enough to provide the same height difference as there was before the bump was hit. Thus, when the wheel gets to the bottom of the bump on the backside, the springs/shocks are still compressed somewhat and that causes either the bumper to scrape the ground past the bump or the underside of the car behind the wheel to scrape the top of the bump.

Also, hitting the bump produces both a vertical force and a horizontal force on the wheel. That horizontal force is in the opposite direction of motion and slows the wheel and chassis down. The body and bumper are connected to the chassis via the springs and shocks and they want to keep going due to momentum. The momentum is centered at the center of gravity of the body/bumper, maybe 2 feet off the ground while the force on the wheel is in the opposite direction at ground level. That creates a 'moment', or 'couple' that causes the body to pitch down in front similar to the way it does when you slam on the brakes. The bumper is out front and so it moves downward the most when the car pitchs down. Also, some people slam on the brakes at a bump which adds to the effect. You can see this effect by slamming on the brakes at slow speed and watching the front of the car pitch downward. So slow down before you get to the bump and then let off the brakes before you get to it. If you're going too fast when you get to the bump, trying to slow down will probably make it worse.

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