Question

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

Answer 1

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