001 (part 1 of 3) 10.0 points In deep space (no gravity), the bolt (arrow) of a crossbow accelerates at 139 m/s 2 and attains a speed of 127 m/s when it leaves the bow. For how long is it accelerated? Answer in units of s
002 (part 2 of 3) 10.0 points What speed will the bolt have attained 3.5 s after leaving the crossbow? Answer in units of m/s
003 (part 3 of 3) 10.0 points How far will the bolt have traveled during the 3.5 s? Answer in units of m
+33665 001.
Use v = u + at (assuming constant acceleration applied by the bowstring).
127 = 0 + 139t
t = 127/139 s
$${\mathtt{t}} = {\frac{{\mathtt{127}}}{{\mathtt{139}}}} \Rightarrow {\mathtt{t}} = {\mathtt{0.913\: \!669\: \!064\: \!748\: \!201\: \!4}}$$
t ≈ 0.9137 s
002.
Since it doesn't accelerate after it leaves the bow and there is nothing to retard it in space, it continues with the same speed: v = 127 m/s.
003.
distance = speed*time
distance = 127*3.5 m
$${\mathtt{distance}} = {\mathtt{127}}{\mathtt{\,\times\,}}{\mathtt{3.5}} \Rightarrow {\mathtt{distance}} = {\mathtt{444.5}}$$
distance = 444.5m
.
+33665 001.
Use v = u + at (assuming constant acceleration applied by the bowstring).
127 = 0 + 139t
t = 127/139 s
$${\mathtt{t}} = {\frac{{\mathtt{127}}}{{\mathtt{139}}}} \Rightarrow {\mathtt{t}} = {\mathtt{0.913\: \!669\: \!064\: \!748\: \!201\: \!4}}$$
t ≈ 0.9137 s
002.
Since it doesn't accelerate after it leaves the bow and there is nothing to retard it in space, it continues with the same speed: v = 127 m/s.
003.
distance = speed*time
distance = 127*3.5 m
$${\mathtt{distance}} = {\mathtt{127}}{\mathtt{\,\times\,}}{\mathtt{3.5}} \Rightarrow {\mathtt{distance}} = {\mathtt{444.5}}$$
distance = 444.5m
.
