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. 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 Z is 457m, find:
(I) the speed of the car at Y;
(ii) the distance from X to Y
(I) To find the speed of the car at Y, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (speed of the car at Y)
u = initial velocity (speed of the car at X = 5 m/s)
a = acceleration (2 m/s^2)
s = distance from X to Y
Initially, v = 5 m/s (speed at X)
Using the equation of motion:
v^2 = u^2 + 2as
v^2 = 5^2 + 2*2*s
v^2 = 25 + 4s
At Y, the speed of the car is v m/s. So, v is the speed of the car at Y.
(ii) To find the distance from X to Y, we can use the equations of motion:
s = ut + 0.5at^2
where:
s = distance
u = initial velocity (speed at X = 5 m/s)
a = acceleration (2 m/s^2)
t = time taken to travel from X to Y
Since the car is accelerating at a constant rate of 2 m/s^2, we can use the equation:
v = u + at
Using the equation v = u + at, we can find the speed of the car at Y, which is the velocity before retardation starts.
v = u + at
v = 5 + 2t (speed at Y before retardation)
After finding the time taken to reach Y, we can substitute the value of t in the equation v^2 = 25 + 4s to find the speed at Y and then calculate the distance from X to Y.