The leader in a bicycle race passes your viewing position at t=0 with a speed of 30 mph. Ten seconds later, the next racer goes by at 40 mph. Assuming their speeds are constant, where will the second cyclist catch up to the first? (Give this as a distance from your position)
I understnad that bike 1 passes at t=o and 10 sec later bike 2 passes. I just don't know what formula to use to help me figure out the problem
First I would suggest converting mph to feet/second for both riders.
When the second rider passes you, the first rider has been riding for 10 seconds past you. Find that distance (10 seconds * first rider's speed in feet/second). The second rider is catching up to the first rider at a speed of 10 mph (convert to ft/sec). Find how many seconds it will take for the second rider to catch the first rider. Multiply this time by the ft/second of the second rider to find how far he (and the first rider) have gone past your position when the second rider cathes the first rider.
First convert the speeds to convenient units of distance per second. I choose ft/s. You can convert to other distances later.
V1 = 30 mph = 44.00 ft/s
V2 = 40 mph = 58.67 ft/s
The distance of the first biker from the viewer is
X1 = 30 t
and the distance if the second biker (after 10 sec) is
X2 = 40 (t-10)
(t will be in seconds, measured from when the first biker goes by)
Set X1 = X2 and solve for t, and then solve for either X.
44 t = 58.67 (t-10) = 58.67 t - 586.7
14.67 t = 586.7
t = 40.0 s
Distance = 40 V1 = 1760 ft = 1/3 mile
To solve this problem, you can use the formula for calculating distance, speed, and time.
First, let's find the time it takes for the second cyclist to catch up to the first cyclist.
Let t be the time it takes for the second cyclist to catch up to the first cyclist.
The leader, the first cyclist, has been cycling for t seconds when the second cyclist passes you at t = 10 seconds.
So the first cyclist has been traveling for t + 10 seconds when the second cyclist catches up.
Now, let's calculate the distance traveled by each cyclist:
The distance traveled by the first cyclist in t + 10 seconds is:
Distance1 = (t + 10) * 30 mph
The distance traveled by the second cyclist in t seconds is:
Distance2 = t * 40 mph
Since the second cyclist catches up to the first cyclist, the distances traveled by both cyclists are equal:
Distance1 = Distance2
Substituting the values:
(t + 10) * 30 mph = t * 40 mph
Now, let's solve this equation to find the value of t:
30t + 300 = 40t
Subtract 30t from both sides:
300 = 10t
Divide both sides by 10:
t = 30 seconds
So, it will take the second cyclist 30 seconds to catch up to the first cyclist.
Now, let's calculate the distance from your position where the second cyclist catches up to the first.
The distance traveled by the first cyclist after t seconds would be:
Distance1 = (t + 10) * 30 mph = (30 + 10) * 30 mph = 40 * 30 mph = 1200 feet
Therefore, the second cyclist will catch up to the first cyclist 1200 feet from your position.