Question
Sitting besides a friend on a park bench, you grab her hat and start running in a straight line away from her. Over the first 15.0, you accelerate at 1 m/s^2 up to your maximum running speed. You then continue at your maximum running speed for 15 s more before your friend catches you. Calculate how far from the bench did you get before being caught and how long did it took your friend to catch up with you.
Answers
d1 = 0.5a*t^2 = 15 m.
0.5*t^2 = 15
t^2 = 30
T1 = 5.477 s.
V = a*t = 1m/s^2 * 5.477s = 5.477 m/s.
d2 = V * t = 5.477m/s * 15s = 82.2 m.
D = d1 + d2 = 15 + 82.2 = 97.2 m. = Distance at which he was caught.
T = T1 + T2 = 5.477 + 15 = 20.48 s To
catch up.
0.5*t^2 = 15
t^2 = 30
T1 = 5.477 s.
V = a*t = 1m/s^2 * 5.477s = 5.477 m/s.
d2 = V * t = 5.477m/s * 15s = 82.2 m.
D = d1 + d2 = 15 + 82.2 = 97.2 m. = Distance at which he was caught.
T = T1 + T2 = 5.477 + 15 = 20.48 s To
catch up.
v2 = sqrt((v1)^2 +2a(x1)) = 5.48m/s
x2 = x1 + (v2 * t2) = 97m
dt = t1 + t2
so, t1 = (v1 - v2)/a = 5.48 s
x2 = x1 + (v2 * t2) = 97m
dt = t1 + t2
so, t1 = (v1 - v2)/a = 5.48 s
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