Acellerating car vs constant speed car

[Lars Krogh-Stea]

Hi, I've been dsicussing this with som friends, and would be thankful if someone took the time to explain it to me :)

The scenario:
Car A travels at constant velocity (80km/h) along the road and hits a brick wall. Car B acellerates upward towards 100km/h, and at excactly the moment it passes 80km/h it hits a brick wall. Which car hits the brick wall with the most force/takes most damage?

If you'll include formulas and calculation, I'' be even more thankful :)

Best Regards, Lars Krogh-Stea

 

[BvU]

The speed $v$ at the time of crashing is the determining quantity. The kinetic energy is ${1\over 2}mv^2$ and that has to be dissipated: converted into deformation of steel and what have you.

 

[A.T.]

Car A travels at constant velocity (80km/h) along the road and hits a brick wall. Car B acellerates upward towards 100km/h, and at excactly the moment it passes 80km/h it hits a brick wall. Which car hits the brick wall with the most force/takes most damage?

For normal cars the force from the engine is much smaller than the collision force at that speed, so there won’t be much difference. But you could construct scenarios where it matters more (very soft car at very low speed with a very strong engine that keeps applying a force during the collision).

 

[Lars Krogh-Stea]

For normal cars the force from the engine is much smaller than the collision force at that speed, so there won’t be much difference. But you could construct scenarios where it matters more (very soft car at very low speed with a very strong engine that keeps applying a force during the collision).

In my example the driver disengages the clutch at the moment of impact, I should have mentioned that.. If there wasn't an impact my believe is that the car would reach maybe 85 or 90 km/h before it starts to slow down as the car doesn't imediately start to slow down, but that the velocity rather follows a trajectory determined by accelleration and resistance.. Doesn't that potential for extra velocity translate into to something at impact? I would believe that the input amount of energy is grater to set a car in motion that has a "velocity trajectory" that reaches 85km/h than 80km/h.. But then again, I'm open to the fact that I still don't understand this.. ;)

 

[A.T.]

In my example the driver disengages the clutch at the moment of impact,

Then both impacts are the same.

If there wasn't an impact my believe is that the car would reach maybe 85 or 90 km/h...

No, it wouldn't

...before it starts to slow down as the car doesn't imediately start to slow down,

Yes it does.

 

[Lars Krogh-Stea]

I get that the accelleration starts to slow down at the time you disengage the clutch, but not that it goes instantly to zero.. I will test this in my car after work, minus the brick wall ;) Thanks for the answers :)

 

[BvU]

Acceleration goes to zero immediately (in fact air & other friction even makes it negative). You feel it: your body and the seat have adjusted to the acceleration -- if it stops you feel pushed forward (especially if you are not the driver).
Test carefully, preferably where there is no other traffic...

 

[A.T.]

I get that the accelleration starts to slow down at the time you disengage the clutch,

It goes below zero due to drag and rolling resistance. Ingoring those you get constant speed in both cases, so still the same impact.

but not that it goes instantly to zero..

Disengaging the clutch is not instantaneous. But at the moment the clutch is completely disengaged (assuming an ideal clutch), the acceleration is zero or less.

 

[Lars Krogh-Stea]

I get it now... If a motorcycle is accellerating off an upward ramp, the highest speed is on the edge of the ramp, even though the bike follows a trajectory. Seems quite logic now that you explained it to me :) Thank you!

 

[A.T.]

I get it now... If a motorcycle is accellerating off an upward ramp, the highest speed is on the edge of the ramp, even though the bike follows a trajectory.

Yes, velocity is different from position. When you throw a ball upwards, the vertical position keeps increasing for a while, but the vertical velocity starts decreasing as soon it leaves your hand.