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

Answer 1

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.

Answer 2

5+2a=0 5=-2a a=-2.5. To find b #-7/2b=0 b= -7

Answer 3

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