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:
Force required to accelerate = 794.44 N
Explanation:
Force required = Mass of horse x Acceleration of horse
Mass of horse and rider, m= 572 kg
Acceleration of horse and rider, a = 5 kph per second

Force required = ma
= 572 x 1.39 = 794.44 N
Force required to accelerate = 794.44 N
Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]
The lateral displacement is I don’t know tbh I think 16.8