Two ice skaters, frank and mary, are initially facing each other and separated by 2000 m on a smooth frozen lake. mary begins skating towards frank at t = 0 s with an acceleration of 0.75 m/s^2. after she reaches a speed of 12 m/s, she continues skating at a constant 12 m/s. three seconds after mary begins to skate, frank starts skating. he accelerates at 1.1 m/s^2 for 9 s and then skates at a constant speed. a) when do the skaters pass each other? b) how far from mary’s starting position is the point at which the skaters pass each other?

a) 99.1 s

b) 1093.2 m

Explanation:

The equation for speed with constant acceleration is

V(t) = V0 + a * t

Mary starts accelerating with a speed of zero, so

V0 = 0

V(t) = a * t

To reach a speed of 12 m/s

t = V(t) / a

t = 12 / 0.75 = 16 s

The equation for position under constant acceleration is

X(t) = X0 + V0 * t + 1/2 * a * t^2

Mary's starting position is zero

X0 = 0

X(t) = 1/2 * a * t^2

X(16) = 1/2 * 0.75 * 16^2 = 96 m

Frank starts skating 3 seconds after Mary, and he accelerates at 1.1 m/s^2 for 9 s. He will stop accelerating at second 12 (9 + 3).

His position after accelerating will be:

X(12) = X0 + V0 * (t - 3) + 1/2 * a * (t - 3)^2

His initial position is 2000, and his initial speed is zero

x(12) = 2000 - 1/2 1.1 * (12 - 3)^2 = 1955.5 m

Shi speed will be

V(12) = -1.1 * (12 - 3) = -9.9 m/s

From there they will move at constant speed from these positions.  We can consider them as moving at constant speed starting at t0 = 16 and t0 = 12 respectively.

For Mary:

X(t) = X0 + V0 * (t - t0)

X(t) = 96 + 12 * (t - 16)

For Frank:

X(t) = 1955.5 - 9.9 * (t - 12)

Equating these two we can find the time when they meet:

96 + 12 * (t - 16) = 1955.5 - 9.9 * (t - 12)

96 + 12*t - 192 = 1955.5 - 9.9*t + 118.8

21.9*t = 2170.3

t = 2170.3 / 21.9 = 99.1 s

Replacing this time value on either equation we get the position:

X(99.1) = 96 + 12 * (99.1 - 16) = 1093.2 m

for this one it would be d bump/forearm pass