Question

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?

Answer 1

`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`

Answer 2

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