In: Physics
From t = 0 to t = 3.90 min, a man stands still, and from t = 3.90 min to t = 7.80 min, he walks briskly in a straight line at a constant speed of 2.96 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 4.90 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 5.90 min?
Final Answer:-
a) 0.76 m/s
b) 0.013 m/s²
c) 1.52 m/s
d) 0.013 m/s²
For time interval 1.00 min to 4.90 min:-
avei velocity is given by,
Vavg = ∆x/∆t
first need to calculate ∆t,
∆t = 4.90 - 1 min
∆t = 3.90 * 60 = 234 seconds
Now the time man is in motion with velocity v,
t = 4.90 - 3.90
t = 1 * 60 = 60 Seconds
Now will calculate ∆x,
∆x = v * t
∆x = 2.96 * 60 = 177.6 m
Now,
Vavg = 177.6/234
Vavg = 0.76 m/s
Now for average acceleration,
Aavg = ∆V/∆t
Aavg = (2.96 - 0)/234
Aavg = 0.013 m/s²
For time interval 2.00 min to 5.90 min:-
avei velocity is given by,
Vavg = ∆x/∆t
first need to calculate ∆t,
∆t = 5.90 - 2 min
∆t = 3.90 * 60 = 234 seconds
Now the time man is in motion with velocity v,
t = 5.90 - 3.90
t = 2 * 60 = 120 Seconds
Now will calculate ∆x,
∆x = v * t
∆x = 2.96 * 120 = 355.2 m
Now,
Vavg = 355.2/234
Vavg = 1.52 m/s
Now for average acceleration,
Aavg = ∆V/∆t
Aavg = (2.96 - 0)/234
Aavg = 0.013 m/s²