Question

From t = 0 to t = 4.66 min, a man stands still, and from t = 4.66 min to t = 9.32 min, he walks briskly in a straight line at a constant speed of 1.66 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 5.66 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 6.66 min?

Answer 1

Average Velocity is given by:

Vavg = dx/dt = (xf - xi)/(tf - ti)

Average acceleration is given by:

a)avg = dV/dt = (Vf- Vi)/(tf - ti)

Part A.

for this part, time interval = dt = 5.66 - 1.00 = 4.66 min

1 min = 60 sec

dt = 4.66*60 = 279.6 sec

His initial position at time 1.00 min is xi = 0 m

Now see that particle is not moving from t = 0 to t = 4.66 min

dt1 = time interval when man was moving

His final position at time 5.66 min is xf = v*dt1 = 1.66*(5.66 - 4.66)*60 = 99.6 m

So,

Vavg = (99.6 - 0)/279.6 = 0.356 m/sec

Part B.

a)avg = (Vf - Vi)/dt

Vi = 0 m/sec at t = 1.00 min, Since initially man stands still

Vf = 1.66 m/sec at t = 5.66 min, Given that constant speed

a)avg = (1.66 - 0)/279.6 = 0.00594 m/sec^2

Part C.

Now In this case

dt = 6.66 - 2 = 4.66 min = 279.6 sec

xi = 0 m, at time t = 2.00 min

xf = v*dt2 = 1.66*(6.66 - 4.66)*60 = 199.2 m

So,

Vavg = (199.2 - 0)/279.6 = 0.712 m/sec

Part D.

a)avg = (Vf - Vi)/dt

Vi = 0 m/sec at t = 2.00 min

Vf = 1.66 m/sec at t = 6.66 min

a)avg = (1.66 - 0)/279.6 = 0.00594 m/sec^2

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