# 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?

Apr 29, 2018

$13333 m , \mathmr{and} 13 k m$.

#### Explanation:

let me rephrase the question to avoid confusion:
A bicycle moves in a straight line with a constant velocity of$\frac{5 k m}{h}$ for 10 mins and then decelerates at the rate of $\frac{1 k m}{{h}^{2}}$ 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

$\frac{5 k m}{h} \cdot \frac{10}{60} h$ $=$ $\frac{5}{6} k m$

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:

$\frac{\frac{5 k m}{h} \cdot 5 h}{2}$$=$12.5km

now the total distance is: $\frac{5}{6} k m + 12.5 k m = 13 \frac{1}{3} k m = 13333 \frac{1}{3} m \approx 13333 m \approx 13 k m$

Apr 29, 2018

#### 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

$v = u + a \cdot t$

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 $v = \frac{s}{t}$

Rearrange to calculate our ${s}_{1}$

${s}_{1} = v \cdot t = 5 k m \cdot 10 \cancel{\min} \cdot \left(\frac{1 h r}{60 \cancel{\min}}\right) = 0.833 k m$

Stage 2: This distance is ${s}_{2}$.
Use the formula ${v}^{2} = {u}^{2} + 2 \cdot a \cdot s$

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).

${v}^{2} = {u}^{2} + 2 \cdot a \cdot s$

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

Solve for ${s}_{2}$

${s}_{2} = \left(25 \left({\text{km"/"hr")^2)/(2*(1 "km"/"hr}}^{2}\right)\right) = 12.5 k m$

Total distance: $0.833 k m + 12.5 k m = 13.33 k m \cong 13 k m$

I hope this helps,
Steve