If the track has circumference c miles, then and the car's speeds are x and y mi/s, then since time = distance/speed,
<span>In the same direction, the slow car has a "lead" of 1 mile which the faster car has to make up. </span>
<span>1/(y-x) = 120 </span>
<span>1/(x+y) = 30 </span>
<span>120(y-x) = 30(x+y) </span>
<span>120y-120x = 30x+30y </span>
<span>90y = 150x </span>
<span>y = 5/3 x </span>
<span>1/(x + 5/3 x) = 30 </span>
<span>8/3 x * 30 = 1 </span>
<span>80x = 1 </span>
<span>x = 1/80 mi/sec = 45 mph </span>
<span>y = 75 mph</span>
Frequency=1/T, where T is the time period.
Reciprocal of frequency = T or it can be written as T = 1/f, where f is frequency.
Calculate the velocity:
v = x/t
v = velocity, x = distance, t = time
Given values:
x = 50m, t = 2.6s
Plug in and solve for v:
v = 50/2.6
v = 19m/s