Question

During a head-on collision, a 22 kg child in the front seat of a car accelerates from 13.7 m/s (≈ 30 miles/hour) to 0 m/s.
Each part below suggests different ways (over distance & time) to stop the kid.
The driver of the car holds out his arm to keep his child (who is not wearing a seat belt) from smashing into the dashboard.
If he can stop the kid, he'll do it by exerting a force over a distance of about 1 m.
(a) What is the acceleration of the child during this stopping motion?
(b) What is the magnitude of the force his arm must exert on the child?

Answer 1

Cont.
The father can't really handle this force, so the dashboard does the stopping.
It will do so by exerting a force over a distance of about 0.049 m (the radius of the kid's head).
(c) What will the child's acceleration be in this case?
(d) What force will the dashboard exert on the child?
Suppose, instead of the scenarios above, that the kid was wearing a seatbelt.
The belt has some 'give' to it, so the force required to stop the kid is exerted over a distance of 0.29 m.
(e) What force will the belt exert on the child?

Answer 2

From the constant acceleration equations I gave you earlier, \[a = {v^2 - v_o^2 \over 2 (x-x_o)}\]

Answer 3

a) the acceleration of the child should be equivalent to mass x accel. which would be 22kg x 13.7 m/s = 301.4 kg m/s.

b) assuming the driver has to stop a force equil to 22 kg accelerating at 13.7 m/s in a forward direction ( due to head on collision) he must exert a magnitude equal to the acceleration and mass of the child, which is 301.4 kg m/s.

the childs acceleration would be the same as in the first case, 301.4 kg m/s.

d) the force exerted from the dashboard to the child can be solved with Force = mass (change in velocity / change in time) giving 22kg ([13.1-0 / x), giving 301.4 /x.

x being the change in time from the childs original position to the dashboard.

not quite sure about the last part.....sorry, hope i helped somewhat, still learning physics on my own, so have someone double check me on all this.

Answer 4

In the interest of helping both of you. I'll run through this.

Part a) From the equation above, \[a = {13.7 \over 2(1)} = 6.85\]
Part b) Force = mass x acceleration. \[F_{arm} = m_{child}a = 22*6.85 = 15.07 [N]\]Repeat using different distance value in acceleration formula. Then use that new acceleration value to solve for force.

Answer 5

oh wow, seems i was wwwwway off in my own world then hahaha.

Answer 6

You got a little confused between FORCE (mass x acceleration) and acceleration itself.