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Solutions |
| 1. |
| (a) 100 m = v1t. Find the v1 = 16.7 m/s |
| (b) -3 m/s2 = (v3
- v2)/(t3 - 6 sec) = (v3 - v1)/(t3 - 6 sec) =(v3 - 16.7 m/s)/(t3 - 6 sec).
Find t3 = 10.23 sec |
| (c) v3 = 4 m/s. v2 = 16.7
m/s. t3 - t2 = 4.23 sec. v32 = v22
+ 2(-3 m/s2)(x3 - x2) ; find x3 - x2 = 43.8 m , or use 43.8 m = x3 - x2 = v2(t3 - t2) + (1/2)(-3 m/s2) (t3 - t2)2 |
| (d) distance = (4.0 m/s) (18 - t3) = 31.08 m |
| (e) a = change in
velocity/change in time = (0 - 4 m/s)/10 sec = -0.4 m/s2 .
Then use displacement = (4.0 m/s)(10 sec) +(1/2)(-0.4 m/s2)(10 sec)2 , or use 02 = 42 + 2(-0.4)(displacement) so (displacement ) = 20 m using either method. |
| (f) see class notes |
| (g) 100m + 43.8m + 31.08m + 20m = 195m |
| (g) 195m/28sec = 7 m/sec |
| 2. |
| (a) Vtop2 = 02 = v12 + 2 (-9.8 m/s2) D so D = 46 m, where v1 = 30.00 m/s |
| (b) Vf 2
= (30)2 + 2 (-9.8 m/s2) (-45 m) so |Vf | =
42.2 m or Vf 2 = (0)2 + 2 (-9.8 m/s2) (-46 m - 45 m) = 2 (-9.8 m/s2) (-96 m) so so |Vf | = 42.2 m |
| (c) Vff 2 = 12 = (-42.2 m)2 + 2 (30 m/s2) (-H) so H = 30 m. |
| 3. (a) Ax = 40cos60 = 20 (b)
Ay = 40sin60 = 34.64 (c) Bx = - 4cos45 = -2.83 (d) By = 4sin45 = 2.83 (e) add components to get Rx = 17.17 (f) add components to get Ry = 37.47 (g) R2 = 2.83 2 + 17.17 2 so R = 41.2. (h) tan(thetaR) = 37.47/17.17 so thetaR = 65 degrees. |
| 4. (a) 1.2 = (1/2)gt2
so t = 0.5 sec. (b) x = Vxt so x = 0.55 m (c) Vx = 1.1 m/s (d) Vy = -gt = -4.9 m/s |