9.
J (0 , -10)
gradient/slope(m) = 2.
We're asked to find K if K is the X- Intercept.
At the x intercept... the corresponding y value is zero. This concept should be pretty clear.
Reason is...
At the Point where the line touches the x axis...If you trace the y value... It will be at point zero(0).
With this Knowledge
We're able to interprete that the points of K is
(x" , 0)
Where x" represents the x coordinate of K
So
J ( 0, - 10)
K ( x" , 0)
m= 2.
Applying the Gradient/Slope Formula
m = y₂ - y₁/x₂ - x₁
x₁ = 0 y₁ = -10
x₂ = x" y₂ = 0
Substituting
2 = 0 - ( -10)/ (x" - 0)
2 = 10/x"
x" = 10/2
x" = 5.
Coordinates of Point K is ( 5 , 0).
THIS IS YOUR ANSWER!.
10.
Total Time Taken = 16secs
10a). We're asked to calculate the rate of speed change for the first 5secs.
This is another way of saying "Calculate Acceleration" in the first 5sec of Motion.
Why?
Acceleration is the rate of change of speed or the rate at which speed changes.
acceleration = v - u/t
where v = Final Speed
u = initial Speed.
Looking at the graph
You see that
Initial speed(u) = 1m/s
Final Speed(v) = 9m/s
t = 5secs
Acceleration = 9 - 1/ 5
a = 8/5
a = 1.6m/s².✅
10b)
We're asked to calc "t" from the graph.
Given that the acceleration at "t" untill 16sec(end of motion) is 2m/s².
This simply means that Constant Acceleration was held from "t" to the end of the Motion.
Do not confuse Constant Acceleration for Constant Velocity.
At Constant Velocity... Acceleration is zero(0)
At Constant Acceleration...Velocity Changes Uniformly or at the same rate each second and this is what we're dealing with in this question.
Taking the same step as we used in calculating acceleration above
Given a= 2m/s²
a = v - u/∆t
Where ∆t means time change or difference.
From "t" to 16
The change in time is 16 - "t"
2 = 21 - 9/( 16 - t)
2 = 12/16-t
16-t = 12/2
16 - t = 6
t = 16 - 6
t = 10secs.✅
Have a great Day!.
