Question

A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His acceleration is constant (4.0 m/s2). At time t =0 he is 5 m east of the signpost and has a velocity of 15 m/s.
1) Find his position and velocity at time t=2 sec.
2) Where is the motorcyclist when his velocity is 25 m/s?
3) If the acceleration is 2.0 m/s2 instead of 4.0 m/s2, where is the cyclist, 5.0 s after he passes the signpost if x0=5.0m and v0x=15m/s?
4) How fast is he moving, 5.0 s after he passes the signpost if x0=5.0m and v0x=15m/s?

Answer 1

Answer:

1) $v_2=23\ m.s^{-1}$              &     $x_2=43\ m$ east of sign post

2) $x'=55\ m$ east of sign post

3) $x_n=205\ m$ east of the signpost.

4) $v_z=35\ m.s^{-1}$

Explanation:

Given:

• position of motorcyclist on entering the city at the signpost, $x_0=0\ m$

• time of observation after being at x=5m east of the signpost, $t_m=0\ s$

• constant acceleration of the on entering the city, $a=4\ m.s^{-2}$

• distance of the motorcyclist moments later after entering, $s_m=5\ m$

• velocity of the motorcyclist moments later after entering, $u_m=15\ m.s^{-1}$

Now the initial velocity on at the sign board:

$u_m^2=u^2+2.a.x_m$

where:

$u=$ initial velocity of entering the city at the signpost

Putting respective values:

$15^2=u^2+2\times 4\times 5$

$u=13.6015\ m.s^{-1}$

1)

Position at time $t_2=2\ s$ sec.:

Using equation of motion,

$x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5$ because it has already covered 5m before that point

$x_2=15\times 2+0.5\times 4\times 2^2+5$

$x_2=43\ m$ east of sign post

Velocity at time $t_2=2\ s$ sec.:

$v_2=u_m+a.t_2$

$v_2=15+4\times 2$

$v_2=23\ m.s^{-1}$

2)

Position when the velocity is $v'=25\ m.s^{-1}$:

using equation of motion,

$v'^2=u_m^2+2.a.x'+5$

$25^2=15^2+2\times 4\times x'+5$

$x'=55\ m$ east of sign post

3)

Given that:

acceleration be, $a_n=2\ m.s^{-2}$

time, $t_n=5\ s$

Position after the new acceleration and the new given time:

using equation of motion,

$x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5$

$x_n=15\times 5+0.5\times 2\times 5^2+5$

$x_n=205\ m$ east of the signpost.

4)

now time of observation, $t_z=5\ s$

$v_z=u_m+a.t_z$

$v_z=15+4\times 5$

$v_z=35\ m.s^{-1}$