A bicycle moves with a constant velocity of 5 km/hr for 10 mins and then decelerates at the rete of 1 km/h till stops. Find the total distance covered by the bicycle?

2 Answers
Apr 29, 2018

13333m, or 13km.

Explanation:

let me rephrase the question to avoid confusion:
A bicycle moves in a straight line with a constant velocity of (5 km)/h for 10 mins and then decelerates at the rate of (1 km)/(h^2) till it stops. Find the total distance covered by the bicycle.

First, to find the distance covered at the first 10 mins, we multiply the bicycle's speed by the time

(5km)/(h)*10/60h = 5/6km

Now we need to find the distance travelled after it starts to decelerate. This is how the speed as a function of time looks like from the moment the bicycle starts to decelerate. (ignore the negative part)
the x-axis represents the time in hours.
the y-axis represents the speed of bicycle in km/hour

graph{y=5-x [-10, 10, -5, 5]}

as shown above, the speed goes from 5km/hour goes down to 0 as time passes.
the distance covered is actually the area of the triangle formed by the diagonal line cutting across the x-axis and y-axis.
Thus the distance traveled here is the surface area of the triangle:

[(5km)/h * 5 h]/2=12.5km#

now the total distance is: 5/6km + 12.5km = 13(1)/(3)km = 13333(1)/(3)m ~~ 13333m ~~13km

Apr 29, 2018

About 13 km.

Explanation:

Deceleration is the rate of decrease of speed. A deceleration of 1 km/h^2 means that after each hour of riding, the speed decreases by 1 km/hr from the speed at the start of that hour. The formula that helps you calculate the new speed is

v = u + a*t

To obtain the total distance, divide the problem into 2 stages:

Stage 1: riding at 5 km/hr for 10 mins
Stage 2: decelerating to a stop at 1 km/h^2

Stage 1: Let's call this distance s_1. Use the formula v = s/t

Rearrange to calculate our s_1

s_1 = v*t = 5 km*10 cancel(min)*((1 hr)/(60 cancel(min))) = 0.833 km

Stage 2: This distance is s_2.
Use the formula v^2 = u^2 + 2*a*s

where v=0 (final velocity is zero),
u=5 km/hr (at the start of stage 2, the speed is 5 km/hr),
and
a = -1 km/hr^2 (the velocity we are using is positive, so forward is the positive direction).

v^2 = u^2 + 2*a*s

0^2 = ((5 "km"/"hr"))^2 + 2*(-1 "km"/"hr"^2)*s_2

Solve for s_2

s_2 = (25 ("km"/"hr")^2)/(2*(1 "km"/"hr"^2)) = 12.5 km

Total distance: 0.833 km + 12.5 km = 13.33 km~= 13 km

I hope this helps,
Steve