Question

Relative velocity of a car

Answer 1

A car is travelling at 13.86m/s. One of the passengers in the car does not have any seatbelts. The car sees a pedestrian, and immediately applies the brakes. The distance the car skids is 10.65m.

Since the passenger wasn't wearing any seatbelts, the passenger flew forward and hit the dashboard. The distance between the her and the dashboard is 0.8m. What is the relative velocity of the girl to the car when she hit the car.

Answer 2

assuming constant (-ve) aceleration of car, using v ^2 = u ^2 + 2ax gives a = (0 - 13.86^2)/(2*10.65) = - 9 m/s/s.

the problem is then ... how long does it take one object (the passenger) travelling at a constant 13.86m/s to catch up with a second object (the dashboard) that has a head start of 0.8m, an initial velocity also of 13.86m/s but an acceleration of -9m/s/s.

to calc time T taken to hit dashboard, use the fact that they end up at the same spot after time = T.

x = x0 + u T + (1/2) a T^2

for passenger
xp = 13.86 T

for dashboard
xd = 0.8 + 13.86 T + (1/2) (-9) T^2

xp = xd ==> 0.8 + (1/2) (-9) T^2 = 0 ==> T = 0.421 s

now calc velocities:

for passenger
vp = 13.86m/s constant (an assumption)
for dashboard
vd = u + at = 13.86 + (-9)(0.421) = 10m.s

rel vel = vp - vd = 3.86m/s.

airbags.

Answer 3

@IrishBoy123, are you saying that the girl hits the dashboard in 0.421s seconds?

Answer 4

I found the relative speed to be 0.5. What about you, @shamim

Answer 5

@shamim @IrishBoy123?

Answer 6

what did you get for T. yes, i reckon the girl hits the dashboard in 0.421s seconds?

Answer 7

Well. I knew there was a distance of 0.8M between the girl and the car. Since they both were at the same speed to begin with, I knew that the car had to travel 0.8M in the opposite direction.
so, v=???, u=13.86, a=9, s=0.8\[
v^2=u^2+2as\]\[v^2=13.86^2+2*-9*0.8\]\[V=13.33\]This is the speed of the car when it moves 0.8 in the opposite direction. Then, I just tok the girls speed and this speed. What is wrong with this solution?

Answer 8

@IrishBoy123?

Answer 9

I think what I did wrong is the fact is the car is still moving forwards. So you can't use this way

Answer 10

No, it would still work, if you consider their relative velocity.

Answer 11

Like, @IrishBoy123, if we consider their relative speedwhen they are both going the same speed

Answer 12

The car is moving behind in relation to the girl. It should travel 0.8M, and deccelerate at 9, right?

Answer 13

remember you have calculated acceleration relative to the ground. you must therefore do the other calculations relative to the ground too.

Answer 14

Yes, but Acceleration of Car to Girl = A Car to Ground - A Girl to Ground, right?

Answer 15

\[C_{AG}=-9-0\]

Answer 16

\[C_{AG}=9\], right?

Answer 17

@IrishBoy123?

Answer 18

relative to ground and on collision, passenger will have moved:
xp = 13.86 * 0.421 = 5.83m

whilst dashboard
xd = 13.86 * 0.421 + (1/2) (-9) (0.421)^2 = 5.03m

Answer 19

So what is wrong with my solution, @IrishBoy123?

Answer 20

You say:
"so, v=???, u=13.86, a=9, s=0.8
v2=u2+2as
v2=13.862+2∗−9∗0.8
V=13.33"

But this only gives you the speed of the car/dashboard after it has travelled 0.8m -- away from the girl. yes?

i m not contesting your logic but your application of this formula is not calculating what you think it should.

Answer 21

Yup. What am I not doing right?

Answer 22

Would I have done it correctly i instead of 13.86, I made it 0 (Relative velocity to the girl), then found v?

Answer 23

Then I get the velocity to be 3.79ms. Would this make sense?

Answer 24

Assume the girl and car have same initial velocity