Constant Acceleration Help?

Question

1. As maria motorsdown the hallway one fine morning, she passes my doorway moving at a speed of 3.0 m/s. A mere 4.2 seconds later she passes Mr. Gonzo's doorway traveling at 5.4 m/s. What is the average acceleration?

2. The data is the position of a rolling ball at various times.

Time (s) Position (m)

0.00 0.480

0.10 0.900

0.20 1.320

0.30 1.740

0.40 2.160

0.50 2.580

0.60 3.000

0.70 3.420

0.80 3.840

0.90 4.260

1.00 4.680

a. How fast was the ball moving at .55 seconds? Justify your answer.

b. What is the average acceleration of the ball?

c. Predict the speed of the ball at 1.2 seconds

Whoever answers this the fastest and the most thorough gets best answer.

Answer 1

1.

For constant acceleration, V(t) = V(0) + a t

V(4.2) = 5.4 = 3 + a (4.2)

a = (5.4 - 3)/4.2

a = 12/21 m/s² or 0.57 m/s²

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2.

Model your rolling ball with constant acceleration.

X(t) = X(0) + V0(t) + 0.5 a t²

Solve with any two data points (or if you want to be very rigorous, do a least squares 2nd order polynomial curve fit).

4.381 = 0.541 + V0(1) + 0.5 a (1)²

0.695 = 0.541 + V0(0.1) + 0.5 a (0.1)²

V0 = 1.284 ; a = 5.111

a. at t=0.55

V(0.55) = V0 + at = 1.284 + 5.111(0.55) = 4.095 m/s

Answer: 4.095 m/s

b. Answer: the average acceleration is 5.111 (see table below, average the fourth column)

time position V (dx/dt) A(dv/dt)

0 0.541

0.1 0.695 1.54

0.2 0.899 2.04 5.000

0.3 1.155 2.56 5.200

0.4 1.463 3.08 5.200

0.5 1.821 3.58 5.000

0.6 2.231 4.1 5.200

0.7 2.691 4.6 5.000

0.8 3.203 5.12 5.200

0.9 3.767 5.64 5.200

1 4.381 6.14 5.000

c. At 1.2 s

V = V0 + at

V = 1.284 + 5.111(1.2)

Answer: V = 7.42 m/s

Answer 2

1. 0.571429 m/s2  (0.571429 X 4.2 = 2.4   3.0+2.4=5.4)

2a The ball is moving at 0.42 m/s all along, even at 0.55s where the positon is 2.79 m

2b constant speed with 0 acceration

2c the speed stays at 0.42 m/s