Answer:
4 seconds
Explanation:
Start with this observation: the distance traveled by car A, plus the distance traveled by car B, plus the original 15 meter gap, equals the 55 meter separation at time t.
dA + dB + 15 = 55
You're given the equation for distance traveled at constant acceleration when the initial velocity is 0:
d = 0.5at²
Given that the acceleration of car A is 2 m/s² and the acceleration of car B is 3 m/s², all you have to do is plug in the expressions and variables and solve for t.
0.5 (2) t² + 0.5 (3) t² + 15 = 55
t² + 1.5 t² = 40
2.5 t² = 40
t² = 16
t = 4
The relationship is an inverse squared relationship.
Answer:
vₓ = 10 m / s, = 10 - 10 t
Explanation:
This is a projectile launching exercise, where the x and y axes are treated independently.
On the x axis. towards the north there is no acceleration, so the speed is constant
vₓ = v₀ₓ
On the vertical y-axis, there is the acceleration of gravity with value (-g) is directed down, the equation is
= - gt
Let's use trigonometry to find the initial velocity component
sin 45 = / v₀
cos 45 = v₀ₓ / v₀
= v₀ sin45
v₀ₓ = v₀ cos45
To find the initial velocity let's use the scope equation
R = v₀² sin 2θ / g
v₀ = √ Rg / sin 2θ
v₀ = √ (20 10 / sin (2 45))
v₀ = 14.14 m / s
With this value we find the initial velocity component
= 14.14 sin 45
= 10 m / s
v₀ₓ = 14.14 cos 45
v₀ₓ = 10 m / s
We get the equations to graph
X axis
vₓ = 10 m / s
In this graph it is a horizontal line
Y Axis
= 10 - 10 t
For this graph let's build a table
t (s) (m / s)
0 10
0.5 5
1.0 0
1.5 -5
2 -10
see attached for both curves