Question

please help with immediacy.

An asteroid is traveling through space at a constant speed of 250 m's relative to a spaceship trailing behind. At the moment the spaceship is 630 m behind, it fires a probe (initially at rest on the spaceship) that accelerates at constant 36 m/s. Treat the spaceship as stationary for this problem. How long will it take for the probe to reach the asteroid? a. b. How far away from the spaceship does the collision between the probe and the asteroid occur? How fast is the probe moving relative to the spaceship when it hits the asteroid? c. 2. Sign the paper

please

Answer 1

Speed of spaceship is $u=0$ ( at rest)

Speed of asteroid with respect to spaceship is $v_A=250\,m/s$

Acceleration of probe is $a=36\,m/s^2$

a) The probe starts from rest ($u=0$) travels a distance $s=630+s'$ and collides with asteroid in a time $t$, in the same time asteroid travels a distance $s'$

Distance traveled by probe is $s=ut+\frac{1}{2}at^2$

$630+s'=0+\frac{1}{2}*36\,t^2$

$630+s'=18\,t^2$ ---(1)

Distance traveled by asteroid is $s'=v_At$

$s'=250t$ ---(2)

from (1) and (2)  $630+250t=18\,t^2$

$18\,t^2-250\,t-630=0$

Solving above quadratic equation, $t=16.067\,s$

Time taken by probe to reach asteroid is $t=16.067\,s\approx 16.1\,s$

b)

$s'=250t=250*16.067=4016.75\,m$

The collision occurs at a distance $s$ from spaceship.

$s=630+s'=630+4016.75=4646.75\,m\approx 4647\,m$

c)

Speed of probe when it collides asteroid is $v_p=u+at=0+36*16.067=578.4\,m/s\approx 578\,m/s$