1) From t = 0 to t = 3.05 min, a man stands still, and from t = 3.05 min to t =
ID: 2270985 • Letter: 1
Question
1) From t = 0 to t = 3.05 min, a man stands still, and from t = 3.05 min to t = 6.10 min, he walks briskly in a straight line at a constant speed of 2.65 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 4.05 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 5.05 min?
2)Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed 15% that of light, which travels at 3.0
Explanation / Answer
1)
a) Vavg = displacement/time
= 2.65*(4.05-3.05)*60/((4.05-1)*60)
= 0.87 m/s
b) aavg = (V2-v1)/(t2-t1)
= (2.65-0)/((4.05-1)*60)
= 0.0145 m/s^2
c) Vavg = displacement/time
= 2.65*(5.05-3.05)*60/((5.05-2)*60)
= 1.74 m/s
d) aavg = (V2-v1)/(t2-t1)
= (2.65-0)/((5.05-2)*60)
= 0.0145 m/s^2
2)
a)u = 0
v = (15/100)*3*10^8 = 4.5*10^7 m/s
v = u + a*t
t = v/a = 4.5*10^7/9.8 = 4.59184*10^6 s
b) s = 0.5*a*t^2 = 4.84*10^13 m
3)
a)s = u*t + 0.5*g*t^2
31 = 11*t + 4.9*t^2
4.9*t^2 + 11*t - 31 = 0
solving abobe equation we get,
t = 1.632 s
b) v = u + a*t = 11 + 9.8*1.632 = 27 m/s
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