XYZ are three points on a straight road.A car passes X with a velocity of 5m/s.It travels from X to Y with a constant acceleration of 2m/s^2 and from Y to Z with constant retardation of 3.5m/s^2.find the velocity of the car as it passes Y and find distance XY

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, it seems like this car is having a bit of a rollercoaster ride on that straight road! Let's see if we can figure out the answers to your questions.

To find the velocity of the car as it passes Y, we can use the equation of motion:
v^2 = u^2 + 2as

In this case, the initial velocity (u) is 5 m/s, the acceleration (a) is 2 m/s^2, and the displacement (s) is unknown. We want to find the final velocity (v) at point Y.

Since the car travels from X to Y, we know that the displacement is the distance XY. Let's call it d.

Plugging in these values, we get:
v^2 = (5 m/s)^2 + 2(2 m/s^2)d

Now, to find the distance XY, we can use another equation of motion:
v^2 = u^2 + 2as

This time, the final velocity (v) is also unknown, and the acceleration (a) is -3.5 m/s^2 (negative because it's retardation). The displacement (s) is still the distance XY.

Plugging in the values, we get:
v^2 = (5 m/s)^2 + 2(-3.5 m/s^2)d

And there you have it! Two equations to find both the velocity as the car passes Y and the distance XY. Now let's solve these equations and see what we get.

To find the velocity of the car as it passes Y, we can use the formula:

v^2 = u^2 + 2as,

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Given that the car passes X with a velocity of 5 m/s, the initial velocity (u) is 5 m/s.

Since the car is traveling from X to Y with a constant acceleration of 2 m/s^2, we need to find the distance XY to use it in the formula.

To find the distance XY, we can use the formula:

s = (v^2 - u^2) / (2a),

where s is the distance, v is the final velocity, u is the initial velocity, and a is the acceleration.

Plugging in the values, we have:

s = (v^2 - 5^2) / (2 * 2),
simplifying, we have:
s = (v^2 - 25) / 4.

Now, let's find the distance XY using the distance formula:

s = ut + (1/2)at^2,

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Since the car is traveling from X to Y with a constant acceleration of 2 m/s^2, we can use this formula with the initial velocity (u) as 5 m/s and the time (t) as unknown.

s = 5t + (1/2)(2)(t^2),
simplifying, we have:
s = 5t + t^2.

Now we have two equations for distance XY. Let's equate them:

(v^2 - 25) / 4 = 5t + t^2.

Simplifying and rearranging the equation, we get:
v^2 = 20t + 4t^2 + 25.

Since the car travels from Y to Z with a constant retardation of 3.5 m/s^2, the final velocity (v) as it passes Y will be the initial velocity (u) for the retardation phase.

Using the formula:
v = u + at,

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given that the retardation is 3.5 m/s^2, we can rewrite the formula as:

v = 5 - 3.5t.

Now, let's solve the equation for v^2:

v^2 = (5 - 3.5t)^2,
v^2 = 25 - 35t + 12.25t^2.

We can equate this equation with the one we found previously:

20t + 4t^2 + 25 = 25 - 35t + 12.25t^2.

Simplifying and rearranging the equation, we get:
16.25t^2 - 55t + 20 = 0.

Now, we can solve this quadratic equation for t. We can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a),

where a = 16.25, b = -55, and c = 20.

Plugging in the values, we can calculate t:

t = (-(-55) ± √((-55)^2 - 4 * 16.25 * 20)) / (2 * 16.25).

Calculating, we get two values for t: t = 1.4 s and t = 2 s.

Now that we have the values for t, we can substitute them back into the equation v = 5 - 3.5t to find the velocities as the car passes Y:

When t = 1.4 s, v = 5 - 3.5 * 1.4 = 5 - 4.9 = 0.1 m/s.
When t = 2 s, v = 5 - 3.5 * 2 = 5 - 7 = -2 m/s.

The velocity cannot be negative, so at t = 2 s, the car is not passing Y.

Therefore, the velocity of the car as it passes Y is 0.1 m/s.

To find the distance XY, we can use the equation s = 5t + t^2 and plug in the value of t = 1.4 s:

s = 5 * 1.4 + (1.4)^2,
s = 7 + 1.96,
s = 8.96 m.

Therefore, the distance XY is 8.96 meters.

To find the velocity of the car as it passes point Y, we can use the equation of motion:

v = u + at

where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time taken.

We know that the initial velocity (u) of the car is 5 m/s, and the acceleration (a) from X to Y is 2 m/s^2. We need to find the time taken (t).

To find the time taken, we can use the equation:

s = ut + (1/2)at^2

where:
s is the distance traveled,
u is the initial velocity,
a is the acceleration, and
t is the time taken.

We know that the distance traveled from X to Y is not given. Let's call it d XY for now.

So, the equation becomes:

d XY = (1/2)(2)t^2

Simplifying, we have:

d XY = t^2

To find the value of t, we need to rearrange the equation:

t = sqrt(d XY)

Now that we know the time taken, we can find the final velocity using the first equation:

v = u + at

Given that the initial velocity (u) is 5 m/s, the acceleration (a) is 2 m/s^2, and the time taken (t) is calculated from the distance XY, we can determine the velocity (v) as the car passes Y.

To find the distance XY, we need more information about the motion of the car from Y to Z.