Question

a cyclist is moving with a constant velocity of 10.8 kmph along a straight road. Another cyclist starts from rest and moves along the same direction with a constant acceleration of 3m/s^2.If both start from the same point,after travelling what distance will they meet again?​

Answer 1

Answer:

6m

Explanation:

Case(i):

• Velocity of first cyclist = 10.8kmph = 10.8*5/18 = 3m/s

• let time taken by him be 't'
• Distance covered by him is Velocity * Time taken = 3t

Case(ii):

• Initial velocity of second cyclist = 0m/s
• Acceleration of second cyclist = 3m/s^2
• let time taken by him be 't'
• Distance covered by him is
• $S=ut+\frac{1}{2}at^2$

• $= 0(t)+\frac{1}{2}3(t)^2=\frac{3t^2}{2}$

As time taken 't' & distance covered will be the same for both cyclists as they meet at the same time at same distance

$=> 3t=\frac{3t^2}{2} \\=> t=t^2/2\\=> 1=t/2\\=>t=2s$

So, time taken is 2s

• We can substitute in distance expression to extract distance covered until they meet

$=>S=3t=3(2)=6m$

Therefore, after travelling 6m distance will they meet again.

Thank you :)