A1 2010 PHYS115-handoutsolutions

A1 2010 - F10-PHYS115-UAE 30 Assignment 1 Solutions 1/4 INTERPRET The problem states that the acceleration of the car is constant so we can use the

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F10-PHYS115-UAE Assignment 1 Solutions 1/4 30. INTERPRET The problem states that the acceleration of the car is constant , so we can use the constant- acceleration equations and techniques developed in this chapter. We’re given initial and final speeds, and the time, and we’re asked to find the distance. DEVELOP Equation 2.9 relates distance to initial and final speed and to time—that’s just what we need. We need the distance traveled during the given time, which would be the difference between x and . o x We also need to be careful of our units—the problem gives us speeds in km/h and time in seconds, so we have to convert so that everything is in the same units. EVALUATE First, convert the speeds to units of m/s. 70 km 1000 m 1 hour 1 hour 1 km 3600 seconds 19.4 m/s.  Similarly, 80 km/h 22.2 m/s. 11 22 () oo o o x xv v t x x v v t      1m m ( ) 19.4 22.2 (6 s) 125 m 2s s o xx     ASSESS These speeds and the time are consistent with passing on a country road. The distance is about right for a passing distance under these circumstances, too; so the answer seems reasonable. 45. INTERPRET This is a one-dimensional problem involving two travel segments. DEVELOP The trip can be divided into two time intervals, 1 t and 2 t with 12 40 min (2 h/3). ttt  The total distance traveled is x x x 25 mi, where 1 x and 2 x are the distances covered in each time interval. EVALUATE During the first time interval, 1 t 15 min (or 0.25 h), with an average speed of 1 v 20 mi/h, the distance traveled is 1 (20 mi/h)(0.25 h) 5 mi t  Therefore, the remaining distance 21 25 mi 5mi 20 mi x   must be covered in 5 40 min 15 min 25 min h 12 tt t   This implies an average speed of 2 2 20 mi 2 5 h/12 48 mi/h. x t v ASSESS The overall average speed was pre-determined to be 25 mi 37.5 mi/h 2 h/3 x v t When you drive slower during the first segment, you make it up by driving faster during the second. In fact, the overall average speed equals the time-weighted average of the average speeds for the two parts of the trip: 12 1 2 1 2 15 min 25 min (20 mi/h) (48 mi/h) 37.5 mi/h 40 min 40 min xx v tv t t t x vv v t t t        58. ( a ) and ( b ) The train goes from velocity 0 110 km/h 30.6 m/s v (positive eastward) at 0 0, t to a stop, 0, v at 1.2 min 72 s. t The constant acceleration was 2 00 ( )/( ) (30.6 m/s)/(72 s) 0.424 m/s .
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This note was uploaded on 10/05/2010 for the course PHYS 115 taught by Professor Rohan during the Fall '06 term at Waterloo.

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A1 2010 - F10-PHYS115-UAE 30 Assignment 1 Solutions 1/4 INTERPRET The problem states that the acceleration of the car is constant so we can use the

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