## College Physics (4th Edition)

(a) The asteroid and the astronaut will be a distance of $18.2~m$ apart. (b) Their relative speed is $3.65~m/s$
(a) We can find the acceleration of the asteroid: $F = ma$ $a = \frac{F}{m}$ $a = \frac{250~N}{40.0~kg}$ $a = 6.25~m/s^2$ We can find the asteroid's speed after the first $0.35~s$: $v_f = v_0+at$ $v_f = 0+(6.25~m/s^2)(0.35~s)$ $v_f = 2.188~m/s$ Note that the asteroid moves at a constant speed after the initial $0.35~s$ acceleration period. We can find the distance the asteroid moves in the next $5.00~s$: $d = v~t = (2.188~m/s)(5.00~s) = 10.9~m$ The asteroid moves a distance of 10.9 meters. By Newton's third law, the asteroid also exerts an opposing force of 250 N on the astronaut. We can find the astronaut's acceleration in the opposite direction: $F = ma$ $a = \frac{F}{m}$ $a = \frac{250~N}{60.0~kg}$ $a = 4.17~m/s^2$ We can find the astronaut's speed after the first $0.35~s$: $v_f = v_0+at$ $v_f = 0+(4.17~m/s^2)(0.35~s)$ $v_f = 1.46~m/s$ Note that the astronaut moves at a constant speed after the initial $0.35~s$ acceleration period. We can find the distance the astronaut moves in the next $5.00~s$: $d = v~t = (1.46~m/s)(5.00~s) = 7.3~m$ The astronaut moves a distance of 7.3 meters in the opposite direction from the asteroid. The asteroid and the astronaut will be a distance of $10.9~m+7.3~m$ apart. The asteroid and the astronaut will be a distance of $18.2~m$ apart. (b) The asteroid has a speed of 2.19 m/s and the astronaut has a speed of 1.46 m/s in the opposite direction. Their relative speed is $2.19~m/s+1.46~m/s$ which is $3.65~m/s$