Question

Two cyclists start together in the same direction from the same place. The first goes with uniform speed of 10km per hour. The second goes at a speed of 8 km per hour in the first hour and increases the speed ½ km each succeeding hour. After how many hours the second cyclist overtake the first if both go non-stop?

Answer 1

for first,
s=vt
in time t, distance(s)=10t
for second,
for time t, s'=8t
seperation= 10t-8t = 2t
for next time t,
s=10t ......................... (1)
s'=8t+1/2(1/2)t^2 ..................(2)
also s'=s+2t
From (1) and (2)
8t+1/2(1/2)t^2 = 10t + 2t
1/4t^2 = 4t
1/4t = 4
t=8
They will meet after 8 hrs

Answer 2

Speed of first cyclist = 10 km/h

Speed of second cyclist:

• In the first hour = 8km/h
• In second hour = (8+1/2) = 17/2 km/h
• In third hour = (17/2 + 1/2) = 9 km/h

The numbers 8,17/2,9....are in AP

• In which a = 8, d = 1/2 and n = x

Let the second cyclist overtake first cyclist in x hours.

∴ Distance covered by the first cyclist in x hours. = distance covered by the second cyclist in x hours = 10x = sum of x numbers in above AP = Sₓ

• ∴ 10x = x/2[2a+(x-1)d]
• ∴ 10x = x/2[2×8+(x-1)1/2]
• ∴ 10x×2/x = 16 + x/2 - 1/2
• ∴ 2×20 = 32+x-1
• ∴ 40 - 31 = x
• x = 9

★ Thus, by 9 hrs the second cyclist overtake the first if both go non-stop.

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