Activity 2: Average Acceleration

1. Explain acceleration in your own words.
Acceleration- measures the change in velocity over the change in time. Any change in velocity is acceleration   (slowing down/speeding up  or changing DIRECTION;  afterall it is a vector not a scalar).

NOTE – it is not change in speed!!
-----------------------------------------------------------------------------

2. If you accelerate due to gravity (free fall), your acceleration is -9.8 m/s2.  In some fields, particularly aviation, this is referred to as 1 g of acceleration. 
A. If you were accelerating at 2.5 g, what would be your average acceleration? 

1x = (9.8)(2.5)
1x = 24.5m/s2   or 24.5 m/s per s   easier to write m/s2
B. If you started from rest, what would be your speed after 10 s?
a = V2-V1/t2-t1
245m/s = V2
Or if I was strick with sig. figs  it would be 2.5x102 m/s

C. What would be your average speed during the acceleration?
Either from your graph   or use   d = (v1 + v2) x time
						    2
=122.5m/s    or 1.2x102 m/s

D. How far would you have travelled in this time? Use average acceleration  so that we can assume acceleration was zero and use the simple formula when a=0.
d= vt
d= (122.5)(10)
d= 1225m   or  1.2x103 m
------------------------------------------------------------------------------------------------------------------

3. A child drops a ball onto the road 28 m in front of a car travelling at 50 km/h.  If the car can decelerate at a rate of 3.5 m/s2, will it stop before hitting the ball?
CONVERT
50km/1h x 1000m/1km x 1h/60min x 5000/3600= 13.89m/s

If a= Δv / Δt    and  Δv=v2 – v1   and Δt = t2 – t1
Then calculate the time to stop  using v2 needs to be zero (you stop)  and t1 is zero, (the time you start)

So Δt = Δv / a     and  Δt = t2 – t1 = t2 – 0 = t2 = t  for simplicity
So t = 0 – (13.89) / -3.5 =  3.97 s   after all deceleration is NEGATIVE (you are braking)
a = V2-V1/t2-t1
-35-t = 0-(-13.89)/0-t = 
-3.5t/-3.5  -13.89/-3.5
t= 3.97 sec

Then distance you will travel when braking is
d=  ( (v1 + v2) x Δt ) /2
d= (v1+ v2/2) (3.968sec)
d= (13.88) (13.968) / 2
d= 27.53 m     You need to keep all the digits in your calculations in order to see that you stop in time.

You can also calculate by area under the graph . . which many of you did.
You can also use the formulas found on  http://zonalandeducation.com/mstm/physics/mechanics/kinematics/EquationsForAcceleratedMotion/EquationsForAcceleratedMotion.htm

Here the site used f for final instead of  2   and o  for initial instead of 1

---------------------------------------------------------------

4. a) first one easy,  your final minus your initial is zero  so zero average acceleration

b)  by calculating the areas under the graphs for the three segments
triangle one : - 37.5 m or use  d=1/2at2

box – constant velocity :  -60m       or use, a= 0  so d=vxt = (-15m/s)(4s) = -60m

triangle two: -22.5 m       or  use    d= v1t  + 1/2at2  where v1=-15m/s , a = -5m/s2 and t= 3s
						
total displacement = -120m