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3 years ago

Question 8 (4 points)

Physics

2 answers:

Paraphin [41]3 years ago

5 0

Answer:

S_d = 5 + 3*t

S_a = 4.2*t

t = 5 / 1.2 = 4.17 s

17.514 m

Explanation:

Given:

- The initial position of Duane s_i = 5

- The speed with which Duane runs v_d = 3 m/s

- The speed with which Albert runs v_a = 4.2 m/s

Find:

a) Write an expression for Duane's position in terms of t and vduane.

b. Write an expression for Albert's position in terms of t and valbert.

c. At what time does Albert catch up to Duane?

d. How far from Albert's starting point are they when they meet?

Solution:

- The position of Duane from start is:

S_d = s_i + v_d*t

S_d = 5 + 3*t

- The position of Albert from start is:

S_a = s_i + v_d*t

S_a = 0 + 4.2*t = 4.2*t

- When Albert catches up-to Duane their positions are equal.

S_a = S_d

5 + 3*t = 4.2*t

1.2*t = 5

t = 5 / 1.2 = 4.17 s

- The distance at which they meet is governed by time t calculated in part above:

S_a = 4.2*(4.17) = 17.514 m

katrin2010 [14]3 years ago

4 0

Answer:

The answer to your question is:

Explanation:

Data

Duane Albert

d = 5 m ; v = 3 m/s v = 4.2 m/s

a) b)

Duane's Albert's

d = 5 + (3)t d = 4.2t

d = 5 + 3t

c) 5 + 3t = 4.2t

4.2t - 3t = 5

1.2t = 5

t = 4.17 s

Duane's

d= 5 + 3(4.17)

d = 17.51 m

Alberts

d = 4.2(4.17)

d = 17.51 m

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