Question

In: Physics

From t = 0 to t = 3.90 min, a man stands still, and from t...

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?

Solutions

Expert Solution

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


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