Respuesta :
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
d)
Duane's
d= 5 + 3(4.17)
d = 17.51 m
Alberts
d = 4.2(4.17)
d = 17.51 m
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