X,Y and Z are three points on a straight road.A car passes X with a speed of 5m/s^-1.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:

Question

(I) the speed of the car at Y;
(ii) the distance from X to Y

Answer 1

(I) Let the speed of the car at point Y be V m/s.

Using the equation of motion v^2 = u^2 + 2as, where:
v = final velocity
u = initial velocity
a = acceleration
s = distance

At point Y:
v^2 = 5^2 + 2(2)s
V^2 = 25 + 4s

At point Z:
0 = V^2 - 2(3.5)(457 - s)
0 = V^2 - 7(457 - s)
0 = V^2 - 3199 + 7s

Substituting V^2 = 25 + 4s into V^2 - 3199 + 7s = 0:
25 + 4s - 3199 + 7s = 0
11s - 3174 = 0
11s = 3174
s = 3174/11
s = 288 m

Therefore, the speed of the car at point Y is:
V^2 = 25 + 4(288)
V^2 = 25 + 1152
V^2 = 1177
V = √1177
V ≈ 34.31 m/s

(ii) The distance from X to Y is 288m.