A motorist travelled from A to B. This is a distance of 142km at an average speed of 60kmhr^(−1) .He spent 5/2hours in B and then returned to A at an average speed of 80kmh^(−1) . a)At what time did the man arrive back at A b)find the average speed for the


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Question Number 45085 by Necxx last updated on 08/Oct/18

$A m o t o r i s t t r a v e l l e d f r o m A t o B .$ $T h i s i s a d i s t a n c e o f 142 k m a t a n$ $a v e r a g e s p e e d o f 60 {k m h r}^{- 1} . H e$ $s p e n t 5 / 2 h o u r s i n B a n d t h e n$ $r e t u r n e d t o A a t a n a v e r a g e s p e e d$ $o f 80 {k m h}^{- 1} .$ $a) A t w h a t t i m e d i d t h e m a n a r r i v e$ $b a c k a t A$ $b) f i n d t h e a v e r a g e s p e e d f o r {t h e}_{}$ $t o t a l j o u r n e y .$
	Answered by MJS last updated on 08/Oct/18

	$journey starts at A at time = 0$ $arrival in B after \frac{142 km}{60 km {hr}^{- 1}} = \frac{71}{30} hr = 2 hr 22 \min$ $time = 2 hr 22 \min$ $5 hr 30 \min in B, start from B at 7 hr 52 \min$ $arrival in A after \frac{142 km}{80 km {hr}^{- 1}} = \frac{71}{40} hr = 1 hr 46 \min 30 s$ $time = 9 hr 38 \min 30 s$ $the average speed (without the 5 : 30 rest) is$ $\frac{2 \times 142 km}{(\frac{71}{30} + \frac{71}{40}) hr} = \frac{480}{7} km {hr}^{- 1} \approx 68 . 57 km {hr}^{- 1}$ $[with the rest included {it}^{'} s \frac{2 \times 142 km}{9 hr 38 \min 30 s} =$ $= \frac{34080}{1157} km {hr}^{- 1} \approx 29 . 46 km {hr}^{- 1}]$
	Commented by Necxx last updated on 09/Oct/18

	$o h . . . T h a n k s$
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