Question

Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts at rest, accelerates with a constant acceleration for 2.7 minutes to a velocity of magnitude 10 m/s. He maintains this velocity for 13.7 minutes and then accelerates, with a constant deceleration, for 4.1 minutes at which time he is at rest. Find the total distance traveled by the rider. Give your answer in kilometers.

Answer 1

Answer:

$s=10.26\ km$

Explanation:

Given

initial velocity of bicycle, $u=0\ m.s^{-1}$

velocity of the bicycle after the first phase of acceleration, $v_1=10\ m.s^{-1}$

duration of first phase of uniform acceleration, $t_1=2.7\ min=162\ s$

duration of second phase of zero acceleration, $t_2=13.7\ min=822\ s$

uniform velocity during the second phase, $v_2=10\ m.s^{-1}$

duration third phase of uniform deceleration, $t_3=4.1\ min=246\ s$

final velocity after the third phase of motion, $v=0\ m.s^{-1}$

• Now we find the acceleration in the first phase of motion:

$a_1=\frac{v_1-u}{t_1}$

$a_1=\frac{10-0}{162} =\frac{10}{162}\ m.s^{-2}$

Now using the equation of motion:

$s_1=u.t_1+0.5a_1.t_1^2$

$s_1=0+0.5\times \frac{10}{162} \times 162^2$

$s_1=810\ m$ is the distance covered in the first phase of motion.

• Distance covered in the third phase of motion:

$s_2=v_2\times t_2$

$s_2=10\times 822$

$s_2=8220\ m$

• Now we find the deceleration in the third phase of motion:

$a_3=\frac{v-v_3}{t_3}$

$a_3=\frac{0-10}{246}$

$a_3=\frac{10}{246}\ m.s^{-1}$

Now using the equation of motion:

$s_3=v_3.t_3+05.\times a_3.t_3^2$

$s_3=10\times 246-0.5\times \frac{10}{246} \times 246^2$

$s_3=1230\ m$ is the distance covered in the third phase of motion.

Hence the total distance covered by the bicycle in the whole incident is:

$s=s_1+s_2+s_3$

$s=810+8220+1230$

$s=10260\ m$

$s=10.26\ km$

Answer 2

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

$v=u+at$

Substitute the values

$10=0+162a=162a$

$a=\frac{10}{162}=0.0617m/s^2$

$s=ut+\frac{1}{2}at^2$

Substitute the values

$s=\frac{1}{2}(0.0617)(162)^2=809.6 m$

$s'=vt'=10\times 822=8220 m$

$a'=\frac{v}{t''}=\frac{10}{246}$

$s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m$

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m