Chapter 2: Q12P (page 52)

A car travels along a straight line at a constant speed of 60.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 30.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
(b) What If? Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip?
(c) What is the average speed for this new trip?

Short Answer

Expert verified

(a) The constant speed of the car during the second distance = 20.0mi/h

(b) The average velocity of the car = 0.00mi/h

Step by step solution

Definition of Average speed

Average speed is the ratio of total distance traveled to total time period is the average speed.

Calculating the time interval of traveling

$(a) T h e c a r t r a v e l s i n t w o d i f f e r e n t w a y s . I t t r a v e l s a d i s \tan c e o f d i n t h e t i m e i n t e r v a l t_{1} a t a c o n s \tan t s p e e d o f v_{1} . T h u s, t i m e i n t e r v a l t a k e n b y t h e c a r i s, t_{1} = \frac{d}{v_{1}} T h e c a r t r a v e l s a n o t h e r d i s \tan c e d w i t h c o n s \tan t s p e e d o f v_{2} i n t h e t i m e i n t e r v a l t_{2} . T h u s, t h e t i m e i n t e r v a l t a k e n b y t h e c a r i s, t_{2} = \frac{d}{v_{2}}$

Calculating the average velocity

$T h e a v e r a g e v e l o c i t y o f t h e e n t i r e m o t i o n i s g i v e n a s f o l l o w s : v_{a v g} = \frac{2 d}{t_{1} + t_{2}} B y s u b s t i t u t i n g \frac{d}{v_{1}} f o r t_{1} a n d \frac{d}{v_{2}} f o r t_{2}, v_{a v g} = \frac{2 d}{\frac{d}{v_{1}} + \frac{d}{v_{2}}} = \frac{2 d}{d (\frac{1}{v_{1}} + \frac{1}{v_{2}})} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}} H e n c e, t h e a v e r a g e v e l o c i t y o f t h e e n t i r e m o t i o n i s v_{a v g} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}} \dots \dots \dots \dots \dots \dots (1)$

Rearrangement of the equation (1) to calculate the value of constant speed of car

$R e a r r a n g e t h e e q u a t i o n v_{a v g} = {\frac{2 v_{1} v_{2}}{v_{1} + v_{2}}}_{{for}^{v_{2}}} v_{a v g} (v_{1} + v_{2}) = 2 v_{1} v_{2} v_{a v g} v_{1} + v_{2} = 2 v_{1} v_{2} v_{2} = \frac{v_{a v g} v_{1}}{2 v_{1} - v_{a v g}} B y s u b s t i t u t i n g 30.0 m i / h f o r v_{a v g} a n d 60.0 m i / h f o r v_{1} v_{2} = \frac{(30.0 m i / h) (60.0 m i / h)}{(2) (60.0 m i / h) - (30.0 m i / h)} = 20.0 m i / h T h e r e f o r e, t h e c o n s \tan t s p e e d o f t h e c a r d u r i n g t h e s e c o n d d i s \tan c e = 20.0 m i / h .$

Definition and calculation of average velocity

(b)

The ratio of total displacement to whole time period is the average velocity.

The average displacement for the entire motion is given as follows:

$∆ x = d + (- d) = 0$

Therefore, the average velocity of the car in this case = 0.0mi/h

Calculation of new average velocity by computing the values

$(c) I t i s g i v e n t h a t t h e t o t a l d i s \tan c e t r a v e l e d i s 2 d a n d t h e c a r r e t u r n s t o t h e h o m e w i t h t h e s a m e s p e e d t h a t f o u n d i n p a r t (a) . S o, v_{a v g} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}} O n s u b s t i t u t i n g 20.0 m i / h f o r v_{2} a n d 60 m i / h f o r v_{1}, v_{a v g} = \frac{(2) (60.0 m i / h) (20.0 m i / h)}{60.0 m i / h + 20.0 m i / h} = 30.0 m i / h T h e r e f o r e, t h e a v e r a g e v e l o c i t y o f t h e n e w t r i p i s 30.0 m i / h .$