Two cyclists start at the same time from a particular place. One cyclist has a speed 1km/hr faster than the second cyclist, The faster cyclist goes in the north direction and the slow cyclist goes in the east direction. If after one hour, they are √221 km apart. Find the speed of the faster cyclist.a
.
.
.
help me please
Answers
Answer:
11 Km/hr
Step-by-step explanation:
Let speed of the slower bicycler be v Km/hr
Then speed of faster bicycler is (v+1) Km/hr {given in question}.
Faster bicycler travels north and slower one travels east i.e they travel perpendicular to each other.
In one hour the faster bicycler travels a distance of (v+1) Km and the slower one travels v Km (since v=s/t).
Observe the picture provided, you can see a right angled triangle.
Therefore by pythagoras theorem distance between the two bicyclers is,
√( (v)² +(v+1)²) . But is given as √221
Therefore,
√(v²+(v+1)²) = √221
v² + (v+1)² = 221
v² + v² + 1 + 2v = 221 ( since (a+b)²=a²+b²+2ab)
2v²+2v-220=0
v²+v-110=0
v²+11v-10v-110=0
v(v+11) -10(v+11) =0
(v-10)(v+11) = 0
∴v=10 or -11
Since we are aware of the directions speed cannot be negative,
Therefore v = 10 Km/hr
Therefore speed of faster bicycler(v+1)=11 km/hr
