How ya doing today sir I’m sure not too good cuz u ain’t gettin the answer sry
Answer:
Explanation:
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1
For David u = 28.0 m/s where 'a' is set to nought
S = ut
S = 28t.......2
For Tina consider equation 1
Where acceleration = 2.90m/s^2 and u is set at nought
S = 1/2×2.90 m/s×t^2.......3
Equate 2 and 3
28t = 1.45t^2
Divide through by t
28 = 1.45t
t = 28/1.45
t = 19.31seconds
Now put the value of t into equation 3
S = 1/2×2.90 m/s×t^2.......3
= 1.45×20×20
= 580m
Tina must have driven 580meters before passing David
Considering the equation of linear motion : V^2 = U^2+2as
Where u is set at nought
V^2 = 2as
V^2 = 2×2.9×580
V^2 = 3364
V = √3364
V = 58m/s
Her speed will be 58m/s
Answer: 20.2 m/s
Explanation:
From the question above, we have the following data;
M1 = 800kg
M2 = 1200kg
V1 = 13m/s
V2 = 25m/s
U (common velocity) =?
M1V1 + M2V2 = (M1 + M2). U
(800*13) + (1200*25) = (800+1200) * U
10400 + 30000 = 2000u
40400 = 2000u
U = 40400 / 2000
U = 20.2 m/s
