A downhill skier crosses the finishing line at a speed of 30
1
ms
−
and immediately starts to
decelerate at 10 m s
−
2
. There is a barrier 50 metres beyond the finishing line.
(a) Find an expression for the skier’s speed when she is
V
metres beyond the finishing
line.
(b) How fast is she travelling when she is 40 metres beyond the finishing line?
(c) How far short of the barrier does she come to a stop?
(a) To find an expression for the skier's speed when she is V meters beyond the finishing line, we need to use the equation for motion with constant acceleration. The equation is:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
Initially, the skier's speed is 30 m/s with no displacement (0 meters). The skier then decelerates at 10 m/s^2. Let's consider the direction of motion as positive.
Applying the equation, we have:
v^2 = (30 m/s)^2 + 2*(-10 m/s^2)*s
Simplifying:
v^2 = 900 m^2/s^2 - 20 m/s^2 * s
Taking the square root:
v = √(900 m^2/s^2 - 20 m/s^2 * s)
Therefore, the expression for the skier's speed when she is V meters beyond the finishing line is:
V = √(900 m^2/s^2 - 20 m/s^2 * s)
(b) To find how fast she is traveling when she is 40 meters beyond the finishing line, we can substitute V = 40 into the expression from part (a):
v = √(900 m^2/s^2 - 20 m/s^2 * 40)
Simplifying:
v = √(900 m^2/s^2 - 800 m^2/s^2)
v = √100 m^2/s^2
v = 10 m/s
Therefore, she is traveling at a speed of 10 m/s when she is 40 meters beyond the finishing line.
(c) To find how far short of the barrier she comes to a stop, we need to find the distance traveled when the skier's speed is zero.
Using the equation v = u + at, where v = 0:
0 = 30 m/s + (-10 m/s^2) * t
Solving for t:
t = 30 m/s / 10 m/s^2
t = 3 s
To find the distance traveled, we use the equation:
s = ut + 0.5 * a * t^2
s = 30 m/s * 3 s + 0.5 * (-10 m/s^2) * (3 s)^2
Simplifying:
s = 90 m - 45 m
s = 45 m
Therefore, she comes to a stop 45 meters short of the barrier.
To solve this problem, we can use the equation of motion for deceleration:
v^2 = u^2 - 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
(a) To find an expression for the skier's speed when she is V meters beyond the finishing line, we need to find the displacement s in terms of V. The skier's initial velocity is 30.1 m/s and the acceleration is -10 m/s^2 since it is decelerating. The displacement will be (V - 50) because the barrier is 50 meters beyond the finishing line.
Using the equation of motion, we can rewrite it as:
v^2 = u^2 - 2as
(v)^2 = (30.1)^2 - 2(-10)(V - 50)
Simplifying the equation, we get:
v^2 = 906.01 + 20V - 1000
v^2 = 20V - 93.99
Taking the square root of both sides, we have:
v = √(20V - 93.99)
So, the expression for the skier's speed when she is V meters beyond the finishing line is √(20V - 93.99).
(b) To find the skier's speed when she is 40 meters beyond the finishing line, substitute V = 40 into the expression we derived in part (a):
v = √(20(40) - 93.99)
v ≈ √(800 - 93.99)
v ≈ √706.01
v ≈ 26.59 m/s
So, the skier is traveling at approximately 26.59 m/s when she is 40 meters beyond the finishing line.
(c) To find how far short of the barrier the skier comes to a stop, we need to find the point where her speed (v) becomes zero. Therefore, we substitute v = 0 into the expression derived in part (a):
0 = √(20V - 93.99)
Squaring both sides:
0 = 20V - 93.99
20V = 93.99
V ≈ 4.70
So, the skier comes to a stop approximately 4.70 meters short of the barrier.
since the speed decreases at 10 m/s every second,
v = 30 - 10t
after t seconds. Looks like it takes 3 seconds to stop. The distance traveled in t seconds is thus
s = 30t - 5t^2
You can use that to figure out (b) and (c)