Question

A bicycle moves with a constant velocity of 5km/h for 10 minutes and then decelerates at rate 1km/h^2 till it stops. Find the total distance covered by bicycle

Answer 1

The total distance covered by bicycle is $\bold{=\frac{40}{3} \mathrm{km}}$.

The bicycle moves at a constant velocity for 10 minutes and then decelerates. The total distance is calculated by forming two equations according to the given data using equation of motion.

Given data:

Initial velocity u = 5 km/hr

Time t = 10 min $=\frac{1}{6} h r$

Deceleration $a=-1\ \mathrm{km} / \mathrm{hr}^{2}$

And so, from the speed formula, we have $\Rightarrow \text { distance }=\text { speed } \times { time }$

Distance $\begin{array}{l}{\mathrm{d} 1=u \times t} \\ {\mathrm{d} 1=5 \times\left(\frac{1}{6}\right)} \\ {\mathrm{d} 1=\frac{5}{6}\ \mathrm{km}}\end{array}$

From equation of motion, we have, $v^{2}=u^{2}+2 a s$

$\begin{aligned} u &=5\ \mathrm{km} / \mathrm{hr} \\ v &=0 \\ a &=-1\ \mathrm{km} / \mathrm{hr}^{2} \\ \mathrm{s} &=\mathrm{d} 2 \end{aligned}$

$\begin{aligned} \mathrm{v}^{2} &=\mathrm{u}^{2}+2 \mathrm{as} \\ 0 &=5 \times 5+(2 \times-1) \times \mathrm{d} 2 \\ 0 &=25-(2 \mathrm{d} 2) \end{aligned}$

$\begin{aligned} 2 d 2 &=25 \\ d 2 &=\frac{25}{2} \end{aligned}$

Therefore, total distance = d1 + d2  

Total distance $\begin{array}{l}{=\left(\frac{5}{6}\right)+\left(\frac{25}{2}\right)} \\ \\{=\frac{(5+75)}{6}} \\ \\{=\frac{80}{6}} \\ \\{=\frac{40}{3} \mathrm{km}}\end{array}$

Answer 2

Explanation:

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