Physics questions involving deceleration, acceleration, motion?

Question

Hi full explanations please if possible. I'm trying to learn here. Thanks

1. A child leaves school and dashes across a crossing in from of a car travelling at 20ms-1. the driver sees him at a distance of 10 m and has an above average reaction time of 0.4s and breaks capable of delivering a deceleration of 20ms-2. Calculate

a) the distance the car moves before the driver reacts.

b) the distance the car moves while braking

c) the total stopping distance

2. A car moves at 20ms-1experience a haed on collision and stops after the bonnet crumples a distance of 0.5m.

Calculate

a.)The time to stop

b,.)The deceleration

3. Repeat if the car is travelling at 40ms-1

Answer 1

Calculations and explanations:

The car is moving at 20 m/s. This is our initial velocity (u). At the end of our observation, the car has stopped, so it's final velocity (v) will be 0.

a) The car moves 20 metres in one second, so in 0.4 of a second, it will move 20*0.4 = 8 metres. Speed = distance/time. So, if we rearrange that, we get distance = speed*time.

b) In kinematic equations, distance is represented by s. Don't know why, it just seems to be. :S :| Anyway, the formula we will use is s=(v^2 - u^2)/2a.

v=0

u=20, 20^2 = 400

a= -20, 2*-20 = -40

So, -400 / -40 = 10 metres.

For the total stopping distance, we'll add the two answers together. 10+8 = 18 metres.

For the second one, we're given distance (0.5), and initial velocity (20). Again, at the end of our observation, the car has stopped, so our final velocity is 0.

We'll use the formula t=s/((u+v)/2).

So this will be 0.5 / (20/2), which is 0.05 seconds.

For the deceleration, we'll use the a=(v-u)/t.

This will be -20/0.05, which is -400 m/s.

There you go, mate!

Answer 2

You need: Dist=rate * time (for zero accel) and

for constant accel, you "need" four equations:

Nomenclature...

d=dist; v=velocity; a=accel; t=time: o=initial; f=final

d=(vo+vf)/2 * t

vf= vo+at

d= 1/2 a t^2 + vo*t + any initial distance

vf^2 = vo^2 + 2ad

Problem 1) in .4 sec, before the driver applies the brakes, the car moves...20 * .4 = 8 meters (dist. car moves before brakes applied)

Now with the brakes on...vf=0, vo=20; a=-20m/s^2 (negative for the deceleration)

Using vf^2 = vo^2 + 2ad,

0 = 400 - 40 d and so...

d=10 meters (distance travelled with vrakes applied)

TOTAL DISTANCE = 18m

Problem 2) Here, d=.5m; vf=0; vo=20m/s

d=(vo+vf)/2 * t gives the time...

.5 = 10 * t, so t=.05 seconds (Ouch!)

vf^2 = vo^2 + 2ad gives us acceleration...

0 = 400 + a and a=-400m/s^2 or about 40g's (Double ouch)

Problem 3) I have shown you the way to do it.  Now you practice this one, OK?  Use mine as a guide.

Take care.