Answer:
Say, the speed of the bicycle before it was decelerated is 245 km/h. Then, John would like to decrease his bicycle's speed by 20%. That means, after the deceleration, he still has 80% of the initial speed.
Let's find out how fast is 80% of the initial speed.
80% * 245 km/h = 196 km/h
So let's say, the terminal speed of the bicycle (after deceleration) is 196 km/h.
Find the difference between the initial speed and terminal speed:
245 km/h - 196 km/h = 49 km/h
Now change the unit of km/h to m/s:
49 km/h = 49 / 3.6 m/s = 13.61 m/s
now determine the time needed to decelerate 20% of the initial speed.
time (t) = difference of velocity (ΔV) / deceleration (d)
t = 13.61 m/s / -3.00 m/s^2
t = 4.537 seconds
Explanation:
smart
Well, to begin with, your first number gang somewhat aglay. The land speed record that John Howard set on his bicycle in 1985 was 152.2 miles per hour, which works out to 68.04 m/s. So I can see where you got the 6 and the 8 from, but your little decimal point snuck over one place when you weren't looking.
I'll use your number to answer the question. If my solution turns out to be wrong, then it's because you copied the number wrong, and you'll have to work it out again with an initial speed of 68.1m/s.
Initial speed = 6.81 m/s
Final speed = 5.44 m/s
Amount of slowing down = 1.37 m/s
Rate at which the brakes slow you down = 3 m/s each second
Time needed to slow down 1.37 m/s = (1.37 m/s) / (3 m/s^2)
That's 0.457 second. (obviously absurd)
If initial speed = 68.1 m/s
Then amount of slowing down = 62.66 m/s
Time needed at -3 m/s^2 = (62.66/3)
That's 20.9 seconds. Much more reasonable.
By the way, John Howard's record was broken 10 yrs later, in 1995 .
