Physx141MPch2sol

Physx141MPch2sol - 2.9. Solve: (a) The time for each of...

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2.9. Solve: (a) The time for each of segment is and The average speed to the house is 1 50 mi/40 mph 5/4 hr t ∆= = 2 50 mi/60 mph 5/6hr. t = 100 mi 48 mph 5/6 hr 5/4 hr = + (b) Julie drives the distance x 1 in time t 1 at 40 mph. She then drives the distance x 2 in time t 2 at 60 mph. She spends the same amount of time at each speed, thus 12 1 2 1 /40 mph /60 mph (2/3) tt x x x ∆=∆⇒∆ =∆ ⇒∆ = 2 x But x 1 + x 2 = 100 miles, so (2/3) x 2 + x 2 = 100 miles. This means x 2 = 60 miles and x 1 = 40 miles. Thus, the times spent at each speed are t 1 = 40 mi/40 mph = 1.00 hr and t 2 = 60 mi/60 mph = 1.00 hr. The total time for her return trip is t 1 + t 2 = 2.00 hr. So, her average speed is 100 mi/2 hr = 50 mph. 2.16. Model: Represent the jet plane as a particle Visualize: Solve: (a) Since we don’t know the time of acceleration, we will use 22 10 2 2 2 1 2( ) (400 m/s) (300 m/s) 8.75 m/s 2 2(4000 m) vv a xx vv a x =+ ⇒= = = (b) The acceleration of the jet is approximately equal to g , the acceleration due to gravity. 2.22. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: Solve: (a) Substituting the known values into 2 1 100 2 yyvt a t = +∆ + ∆ , we get 11 1 10 m 0 m 20 (m/s) ( 9.8 m/s ) 2 tt −=+ + One of the roots of this equation is negative and is not relevant physically. The other root is t 1 = 4.53 s which is the answer to part (b). Using , we obtain vva =+∆ t
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2 1 20(m/s) ( 9.8 m/s )(4.53 s) 24.4 m/s v =+ = (b) The time is 4.53 s.
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Physx141MPch2sol - 2.9. Solve: (a) The time for each of...

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