Friday 5 Nov:

Target Today: Constant Acceleration

  Journal: How can we measure acceleration (Formulas)?

Lab: Acceleration of Carts Today Friday 5 Nov

Test 9, 12, 13, 15, 16, 17  Tuesday 9 Nov 

L# 19: 1- 4, 6-15   Due Wed 10 Nov

L# 20: 1- 5, 7, 9 - 16   Due Friday 12 Nov

L# 21: 1,2, 4 - 6, 9-16   Due Monday 15 Nov

L# 22: 1- 5, 7- 9, 12, 14 - 20   Due Wed 17 Nov

L# 23: 1-4, 8, 13, 15, 16, 18, 20   Due Friday 19 Nov

Review Monday 22 Nov

Test  L# 19 - 23  Tues 23 Nov

. Bridge Contest:

• Due by Friday Dec 17

Tower Contest:

• Due by Friday Jan 7th

Lab: Acceleration of Carts

Question: Can the acceleration of a cart be found using 50g, 100g & 150g forces?

Background: (On force and acceleration)

Hypothesis: Using the forces (0.5N, 1.0N, & 1.5N) ) on the cart & mass system, the acceleration of the cart & mass system will be:

 a)_____  for the 0.5N force;

b) _____  for the 1.0N force;

c) _____  for the 1.5N force;

because ....

Material: (Yours)

Procedures: (Yours)

Data:   (Yours)

Record for each of the 3 forces used:

 1. Final Velocity (V₂) from photo gates

Note: V₂ = .05m/photo-gate time(t₁)

 2. Time (t₂) it took the cart to traveled to the photo-gate.

(see #1 above)

• Use formula to find acceleration:

  acceleration = DV/t  (i.e., V₂-V₁/ t₂ time of cart run)

(Must have a graph of force vs acceleration for)

Conclusion:   (Yours)

1. A hawk can dive @ 70m/s. How long to reach the ground from a height of 200M?

 x = (v)(t) à 200 = 70 t  à  t = 2.86s

2. A torpedo is fired with a constant speed of 30 m/s at a target. After traveling 900m at that speed, the torpedo runs out of fuel and has a negative acceleration of 0.5m/s².

Find:             

a. Time moving at constant speed?

b. Time moving during acceleration period?       

c. Total time to stop from firing pt?

d. Distance moved during acceleration period?

e. Total distance traveled?

2.         a) x = vt à 900 = 30t  à  t = 30s

b. V_f = V_o + at à 0 = 30 + -.5t  à t = 60s

c.     30s + 60s = 90s

d. X = V_ot + ½ at²  à (30x60) +.5(-.5)(60²)

à 1800 + (-900) à  x = 900m

e. Total dist = 900m + 900 = 1800m

3. A golf ball is hit with an initial speed of 100m/s (no air friction). at an angle of 70⁰

            a. How long to the Apex?

            b. How high to the Apex?

            c. Distance Traveled?

a.  

Time to apex:  (depends only on vertical velocity)

         V_y =  (Sin 70)(100) = 94m/s

        V_y =  V_{o +} at   --> 0 = (94) + (-10)t

          à t = 94/10 à 9.4s to apex

                           (18.8 sec total time)

b) dist_y = V_ot + ½ at² 

          dist_y = (94)(9.4)+ .5 (-10)(9.4²)

           dist_y = (884m) + (- 442m)  = 442m

 c. Horz V (no acceleration in x direction)

                 V_x =    V_o     +      at  

       V_x  = (Cos 70)(100) +  0    à 34 m/s

                    dist  = (V_x)(total time)

                              (34m/s)(18.8s)  =   640m