from the same place at 7:00 a.m. started walking in the north at a speed of 5 kilometre per hour after 1 hour be started cycling in the east east adress speed of history kilometre per hour at what time they will be at a distance of 52 km apart from each other
Answers
Answer:
Let t =t= travel time of bicycle
then (t + 1) =(t+1)= Walking time
Distance == Speed \times× Time
This is a Pythagoras problem
a^2 + b^2 = c^2a
2
+b
2
=c
2
Where a = 8ta=8t, bicycle distance
b = 4 (t + 1)b=4(t+1), Walking distance
c = 20c=20, distance apart in tt hrs from 8\ a.m8 a.m.
\therefore (8t)^2 + [4(t + 1)]^2 = (20)^2∴(8t)
2
+[4(t+1)]
2
=(20)
2
64t^2 + 16(t^2 + 1 + 2t) = 40064t
2
+16(t
2
+1+2t)=400
64t^2 + 16t^2 + 32t + 16 = 40064t
2
+16t
2
+32t+16=400
80t^2 + 32 t - 384 = 080t
2
+32t−384=0
16(5t^2 + 2t - 24) = 016(5t
2
+2t−24)=0
5t^2 + 2t - 24 = 05t
2
+2t−24=0
5t^2 + (12 - 10) t - 24 = 05t
2
+(12−10)t−24=0
5t^2 + 12t - 10 t - 24 = 05t
2
+12t−10t−24=0
5t^2 - 10t + 12 t - 2 4 = 05t
2
−10t+12t−24=0
5t(t - 2) + 12(t - 2) = 05t(t−2)+12(t−2)=0
(t -2)(5t + 12) = 0(t−2)(5t+12)=0
t = 2t=2
and t = - \dfrac{12}{15}t=−
15
12
which is not possible.
Positive solution t = 2\ hrst=2 hrs.
At 10\ a.m.10 a.m. they will be 20\ km20 km apart.
Step-by-step explanation: