Respuesta :
Answer:
1) [tex]v_2=23\ m.s^{-1}[/tex] & [tex]x_2=43\ m[/tex] east of sign post
2) [tex]x'=55\ m[/tex] east of sign post
3) [tex]x_n=205\ m[/tex] east of the signpost.
4) [tex]v_z=35\ m.s^{-1}[/tex]
Explanation:
Given:
- position of motorcyclist on entering the city at the signpost, [tex]x_0=0\ m[/tex]
- time of observation after being at x=5m east of the signpost, [tex]t_m=0\ s[/tex]
- constant acceleration of the on entering the city, [tex]a=4\ m.s^{-2}[/tex]
- distance of the motorcyclist moments later after entering, [tex]s_m=5\ m[/tex]
- velocity of the motorcyclist moments later after entering, [tex]u_m=15\ m.s^{-1}[/tex]
Now the initial velocity on at the sign board:
[tex]u_m^2=u^2+2.a.x_m[/tex]
where:
[tex]u=[/tex] initial velocity of entering the city at the signpost
Putting respective values:
[tex]15^2=u^2+2\times 4\times 5[/tex]
[tex]u=13.6015\ m.s^{-1}[/tex]
1)
Position at time [tex]t_2=2\ s[/tex] sec.:
Using equation of motion,
[tex]x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5[/tex] because it has already covered 5m before that point
[tex]x_2=15\times 2+0.5\times 4\times 2^2+5[/tex]
[tex]x_2=43\ m[/tex] east of sign post
Velocity at time [tex]t_2=2\ s[/tex] sec.:
[tex]v_2=u_m+a.t_2[/tex]
[tex]v_2=15+4\times 2[/tex]
[tex]v_2=23\ m.s^{-1}[/tex]
2)
Position when the velocity is [tex]v'=25\ m.s^{-1}[/tex]:
using equation of motion,
[tex]v'^2=u_m^2+2.a.x'+5[/tex]
[tex]25^2=15^2+2\times 4\times x'+5[/tex]
[tex]x'=55\ m[/tex] east of sign post
3)
Given that:
acceleration be, [tex]a_n=2\ m.s^{-2}[/tex]
time, [tex]t_n=5\ s[/tex]
Position after the new acceleration and the new given time:
using equation of motion,
[tex]x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5[/tex]
[tex]x_n=15\times 5+0.5\times 2\times 5^2+5[/tex]
[tex]x_n=205\ m[/tex] east of the signpost.
4)
now time of observation, [tex]t_z=5\ s[/tex]
[tex]v_z=u_m+a.t_z[/tex]
[tex]v_z=15+4\times 5[/tex]
[tex]v_z=35\ m.s^{-1}[/tex]
