Three towns x,y,z are on a straight road and y is midway between x and z, a motorcyclist moving with a uniform acceleration pass through x,y,z, the speed with which the cyclist passes x and z are 20m/s and 40m/s respectively, find the speed with which the motorcyclist passes y

Answer 1

Let the initial speed of the motorcyclist be u m/s, the acceleration be a m/s^2, the distance between x and y be d1, and the distance between y and z be d2.

Given:
Speed at x = 20 m/s
Speed at z = 40 m/s

Using the first equation of motion, we have:
v^2 = u^2 + 2as

At x (initial point):
20^2 = u^2 + 2ad1 ----(1)

At z (final point):
40^2 = u^2 + 2ad2 ----(2)

Since y is the midpoint between x and z:
d1 = d2 = d

From equations (1) and (2), we get:
400 = u^2 + 2ad ----(3)

1600 = u^2 + 4ad ----(4)

Subtracting equation (3) from equation (4), we get:
1200 = 2ad

Since d = (d1 + d2)/2 = 3d/2, we get:
1200 = 2a * 3d/2
a = 400/d ----(5)

Now, when the motorcyclist is at y, the speed v can be found using equation of motion:
v^2 = u^2 + 2ad/2
v^2 = u^2 + ad

Substitute a from equation (5) into the equation above, we get:
v^2 = u^2 + 400
v^2 = 400 + 400
v^2 = 800
v = sqrt(800) = 20sqrt(2)

Therefore, the speed with which the motorcyclist passes y is 20sqrt(2) m/s.