Respuesta :
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
[tex]v=u+at[/tex]
Substitute the values
[tex]10=0+162a=162a[/tex]
[tex]a=\frac{10}{162}=0.0617m/s^2[/tex]
[tex]s=ut+\frac{1}{2}at^2[/tex]
Substitute the values
[tex]s=\frac{1}{2}(0.0617)(162)^2=809.6 m[/tex]
[tex]s'=vt'=10\times 822=8220 m[/tex]
[tex]a'=\frac{v}{t''}=\frac{10}{246}[/tex]
[tex]s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m[/tex]
Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m
Answer:
[tex]s=10.26\ km[/tex]
Explanation:
Given
initial velocity of bicycle, [tex]u=0\ m.s^{-1}[/tex]
velocity of the bicycle after the first phase of acceleration, [tex]v_1=10\ m.s^{-1}[/tex]
duration of first phase of uniform acceleration, [tex]t_1=2.7\ min=162\ s[/tex]
duration of second phase of zero acceleration, [tex]t_2=13.7\ min=822\ s[/tex]
uniform velocity during the second phase, [tex]v_2=10\ m.s^{-1}[/tex]
duration third phase of uniform deceleration, [tex]t_3=4.1\ min=246\ s[/tex]
final velocity after the third phase of motion, [tex]v=0\ m.s^{-1}[/tex]
- Now we find the acceleration in the first phase of motion:
[tex]a_1=\frac{v_1-u}{t_1}[/tex]
[tex]a_1=\frac{10-0}{162} =\frac{10}{162}\ m.s^{-2}[/tex]
Now using the equation of motion:
[tex]s_1=u.t_1+0.5a_1.t_1^2[/tex]
[tex]s_1=0+0.5\times \frac{10}{162} \times 162^2[/tex]
[tex]s_1=810\ m[/tex] is the distance covered in the first phase of motion.
- Distance covered in the third phase of motion:
[tex]s_2=v_2\times t_2[/tex]
[tex]s_2=10\times 822[/tex]
[tex]s_2=8220\ m[/tex]
- Now we find the deceleration in the third phase of motion:
[tex]a_3=\frac{v-v_3}{t_3}[/tex]
[tex]a_3=\frac{0-10}{246}[/tex]
[tex]a_3=\frac{10}{246}\ m.s^{-1}[/tex]
Now using the equation of motion:
[tex]s_3=v_3.t_3+05.\times a_3.t_3^2[/tex]
[tex]s_3=10\times 246-0.5\times \frac{10}{246} \times 246^2[/tex]
[tex]s_3=1230\ m[/tex] is the distance covered in the third phase of motion.
Hence the total distance covered by the bicycle in the whole incident is:
[tex]s=s_1+s_2+s_3[/tex]
[tex]s=810+8220+1230[/tex]
[tex]s=10260\ m[/tex]
[tex]s=10.26\ km[/tex]
