Chapter 2: Q83P (page 58)

In a women’s 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 s to attain their maximum speeds, which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a world record of 10.4 s. (a) What is the acceleration of each sprinter? (b) What are their respective maximum speeds? (c) Which sprinter is ahead at the 6.00-s mark, and by how much? (d) What is the maximum distance by which Healan is behind Laura, and at what time does that occur?

Short Answer

Expert verified

$(a) T h e a c c e l e r a t i o n o f L a u r a a n d H e a l a n i s 5.32 m / s^{2} a n d 3.75 m / s^{2} . (b) T h e m a x i m u m s p e e d o f L a u r a a n d H e a l a n i s 10.64 m / s a n d 11.25 m / s . (c) T h e L a u r a i s a h e a d t o H e a l a n b y a d i s \tan c e o f 2.6 m . (d) T h e H a l e n i s b e h i n d L a u r a b y m a x i m u m d i s \tan c e 4.45 m a t 2.84 s .$

Step by step solution

Identify given data

$T h e t i m e t a k e n b y t h e L a u r a t o a c h i e v e m a x s p e e d i s t_{l} = 2 s T h e t i m e t a k e n b y t h e H e a l a n t o a c h i e v e m a x s p e e d i s t_{h} = 3 s T h e t i m e t o c o m p l e t e r a c e i s T = 10.4 s T h e t i m e f o r t h e l e a d b e t w e e n s p r i n t e r s i s t = 6 s T h e d i s \tan c e f o r t h e r a c e i s d = 100 m$

Determine acceleration of Laura and Haelan

$(a) T h e a c c e l e r a t i o n o f t h e L a u r a i s c a l c u l a t e d a s : d = \frac{1}{2} a_{l} {t^{2}}_{l} + 2 a_{l} (T - t_{l}) S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . 100 m = \frac{1}{2} a_{l} {(2 s)}^{2} + 2 a_{l} (10.4 s - 2 s) a_{l} = 5.32 m / s^{2} T h e a c c e l e r a t i o n o f t h e H a l e n i s c a l c u l a t e d a s : d = \frac{1}{2} a_{h} {t_{h}}^{2} (T - t_{n}) S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . 100 m = \frac{1}{2} a_{h} {(3 s)}^{2} + 2 a_{h} (10.4 s - 3 s) a_{h} = 3.75 m / s^{2} T h e r e f o r e, t h e a c c e l e r a t i o n o f L a u r a a n d H a e l a n i s 5.32 m / s^{2} a n d 3.75 m / s^{2} .$

Determine maximum speed of Laura and Haelan

$(b) T h e m a x i m u m s p e e d o f L a u r a i s g i v e n a s : v_{l} = a_{l} t_{l} S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . v_{l} = (5.32 m / s^{2}) (2 s) v_{l} = 10.64 m / s T h e m a x i m u m s p e e d o f H a l e n i s g i v e n a s : v_{h} = a_{h} t_{h} S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . v_{h} = (3.75 m / s^{2}) (3 s) v_{h} = 11.25 m / s T h e r e f o r e, t h e m a x i m u m s p e e d o f L a u r a a n d H a e l a n i s 10.64 m / s a n d 11.25 m / s .$

Determination of distance between sprinters at 6 s

$(c) T h e d i s \tan c e c o v e r e d b y L a u r a f o r l e a d p o s i t i o n i s g i v e n a s : d = \frac{1}{2} a_{l} {t_{l}}^{2} + v_{l} t S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . d_{1} = \frac{1}{2} (5.32 m / s^{2}) {(2 s)}^{2} + (10.64 m / s) (6 s) d_{1} = 53.2 m T h e d i s \tan c e c o v e r e d b y H a l e n f o r l e a d p o s i t i o n i s g i v e n a s : d_{2} = \frac{1}{2} a_{h} {t_{h}}^{2} + v_{h} t S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . d_{2} = \frac{1}{2} (3.75 m / s^{2}) {(3 s)}^{2} + (11.25 m / s) (3 s) d_{2} = 50.63 m T h e d i f f e r e n c e i n d i s \tan c e o f L a u r a a n d H a l e n i s g i v e n a s : x = d_{1} - d_{2} x = 53.2 - 50.63 x = 2.6 m T h e r e f o r e, t h e L a u r a i s a h e a d t o o f H a l e n b y a d i s \tan c e o f 2.6 m .$

Demonstrate maximum distance by which the Haelan is behind Laura

$(d) T h e t i m e b y w h i c h H a l e n i s b e h i n d L a u r a i s g i v e n a s : v = a_{h} t_{1} S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . 10.64 m / s = (3.75 m / s^{2}) t_{1} t_{1} = 2.84 s T h e m a x i m u m d i s \tan c e b y w h i c h H a l e n i s b e h i n d L a u r a i s g i v e n a s : y = \frac{1}{2} a_{l} {t_{l}}^{2} + (t_{1} - t_{l}) v_{l} - \frac{1}{2} a_{h} {t_{1}}^{2} S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . y = \frac{1}{2} (5.32 m / s^{2}) {(2 s)}^{2} + (10.64 m / s) (2.84 s - 2 s) - \frac{1}{2} (3.75 m / s^{2}) {(2.84 s)}^{2} y = 4.45 m T h e r e f o r e, t h e H a e l a n i s b e h i n d L a u r a b y m a x i m u m d i s \tan c e 4.45 m a t 2.84 s .$

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Short Answer

Step by step solution

Identify given data

Determine acceleration of Laura and Haelan

Determine maximum speed of Laura and Haelan

Determination of distance between sprinters at 6 s

Demonstrate maximum distance by which the Haelan is behind Laura

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