X, Y and Z are three points on a straight road.A car passes X with a speed of 5m/s and travels from X to Y with a constant acceleration of 2m/s^2. The car then moves with a constant retardation of 3.5m/s^2 from Y to Z and comes to rest at Z. If the total distance from X to Y is 475m, find (i) the speed of the car at Y (ii) the distance from X to Y
a1= 2m/s
u1=5m/s
a2=3.5m/s
Total Distance S=475m
If distance X to Y = a
the Formular
v^2 = u^2 + 2as
v^2 = 5^2 + 2x2xa
v^2 = 25 + 4a ---eqn (1)
From Y to Z
Distance = 475 - a
using same formular, v = 0, initial speed from Y to Z is the final speed of X to Y
therefore
v^2 = u^2 + 2as
0^2 = u^2 - 2x3.5 x(475 - a)
u^2 = 3325 - 7a ----eqn (2)
since u^2 = v^2 , then
3325 - 7a = 25 + 4a
3325 - 25 = 7a + 4a
3300/11 = 11a/11
a = 300 means Distance X to Y = 300m
Speed of the car at Y, substitute into equation (1)
v^2 = 25 + 4a
v^2 = 25 + 4x300
v^2 = 25 + 1200
v^2 = 1225
v= square root of 1225
v = 35m/s
Well, well, it seems like our car is having quite the journey on this straight road. Let's dive into the details, shall we?
(i) To find the speed of the car at Y, we can use a handy formula. We know that the car's initial speed at X is 5 m/s, and it has a constant acceleration of 2 m/s^2 from X to Y. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since we're dealing with Y, let's call the time it takes to travel from X to Y as t1. And let's call the speed of the car at Y as vY (creative, right?).
Using the formula, we can say vY = 5 + 2t1. Keep that in mind because we'll come back to it.
(ii) Now, let's tackle the distance from X to Y. We're given that it's 475 m. We can use another handy formula here: s = ut + 1/2at^2. In this case, s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Since we know the car's initial speed at X, which is 5 m/s, and the acceleration it experiences from X to Y, which is 2 m/s^2, we can plug in the values and solve for t1.
475 = 5t1 + 1/2(2)(t1^2)
Now, this isn't too fun to solve, but let's power through it!
Once we find t1, we can plug it back into the equation we found earlier, vY = 5 + 2t1, to determine the speed of the car at Y.
And voila! We've got answers to both (i) and (ii). Not too shabby, huh?
To find the speed of the car at point Y and the exact distance from point X to point Y, we can use the equations of motion.
Let's start by finding the speed of the car at point Y (i).
We are given:
Initial speed (u) = 5 m/s
Acceleration (a) = 2 m/s^2
Distance (s) = 475 m (from X to Y)
We'll use the kinematic equation:
v^2 = u^2 + 2as
First, let's rearrange the equation to solve for v (final speed):
v^2 = u^2 + 2as
v^2 = (5 m/s)^2 + 2 * (2 m/s^2) * (475 m)
v^2 = 25 m^2/s^2 + 2 * 2 * 475 m^2/s^2
v^2 = 25 m^2/s^2 + 4 * 475 m^2/s^2
v^2 = 25 m^2/s^2 + 1900 m^2/s^2
v^2 = 1925 m^2/s^2
Taking the square root of both sides gives us:
v = √(1925 m^2/s^2)
v ≈ 43.92 m/s
Therefore, the speed of the car at point Y is approximately 43.92 m/s.
Now, let's find the distance from point X to point Y (ii).
We can use a different equation of motion:
v^2 = u^2 + 2as
Substituting the given values:
0 = (43.92 m/s)^2 + 2 * (-3.5 m/s^2) * s
Rearranging the equation:
2 * (-3.5 m/s^2) * s = -(43.92 m/s)^2
-7 m/s^2 * s = -(43.92 m/s)^2
s ≈ (43.92 m/s)^2 / (7 m/s^2)
s ≈ 275.6 m^2/s^2 / 7 m/s^2
s ≈ 39.37 m
Therefore, the distance from point X to point Y is approximately 39.37 m.
5+2a=0 5=-2a a=-2.5. To find b #-7/2b=0 b= -7
let the time spent on XY and YZ be a and b, respectively. Then
on XY: s = 5a + a^2; v@Y=5+2a
on YZ: s = (5+2a)b - 7/4 b^2; v@Z=0
Then we have
5a + a^2 + (5+2a)b - 7/4 b^2 = 475
5 + 2a - 7/2 b = 0
Just solve for a and b, and then you can answer the questions.