When a high-speed passenger train traveling at 161 km/h rounds a bend,the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding
and is a distance(D)= 676m ahead.The locomotive is moving at 29.0 km/h.The engineer of the high-speed train immediately applies the brakes.What must be the magnitude of the
resulting constant deceleration if a collision is to be just avoided
Solve
To solve this problem, we can use the equations of motion.
Let:
- v1 be the initial velocity of the high-speed train (161 km/h)
- v2 be the velocity of the locomotive (29.0 km/h)
- D be the initial distance between the train and the locomotive (676 m)
- a be the constant deceleration of the high-speed train
We want to find the magnitude of the resulting constant deceleration (a) to just avoid a collision.
First, let's convert the velocities from km/h to m/s:
v1 = 161 km/h * (1000 m/1 km) * (1 h/3600 s) = 44.7 m/s
v2 = 29.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 8.06 m/s
Now, we can use the equations of motion to solve for the constant deceleration (a).
The equation to calculate the distance covered during deceleration is:
D = (v1^2 - v2^2)/(2a)
Plugging in the given values:
676 m = (44.7^2 - 8.06^2)/(2a)
Simplifying the equation:
676 m = (1998.09 - 64.9636)/(2a)
676 m = 1933.1264/(2a)
Now, solve for a:
a = 1933.1264/(2 * 676 m)
a = 1.43 m/s^2
Therefore, to just avoid a collision, the magnitude of the resulting constant deceleration must be approximately 1.43 m/s^2.
To solve this problem, we need to find the constant deceleration required for the high-speed train to avoid a collision with the locomotive.
We can start by converting the speeds to meters per second (m/s) for consistency.
The speed of the high-speed train is given as 161 km/h. To convert this to m/s, we divide by 3.6 (since 1 km/h is equal to 1/3.6 m/s):
Speed of high-speed train = 161 km/h ÷ 3.6 = 44.7 m/s
The speed of the locomotive is given as 29.0 km/h. Converting this to m/s:
Speed of locomotive = 29.0 km/h ÷ 3.6 = 8.1 m/s
Now, let's consider the time it takes for the high-speed train to cover the distance D and collide with the locomotive. We can use the equation:
D = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the high-speed train is applying brakes immediately, its initial velocity is 44.7 m/s, and the initial velocity of the locomotive is 8.1 m/s.
We can assume the time taken to collide is the same for both the high-speed train and the locomotive. Let's call this time 't'.
For the high-speed train:
D = (44.7 m/s * t) + (0.5 * a * t^2) --- Equation 1
For the locomotive:
D = (8.1 m/s * t) --- Equation 2
Since D is the same in both equations, we can equate Equation 1 and Equation 2:
(44.7 m/s * t) + (0.5 * a * t^2) = (8.1 m/s * t)
Simplifying and rearranging the equation, we get:
0.5 * a * t^2 + 44.7 m/s * t - 8.1 m/s * t = 0
Now, we can solve this quadratic equation for the acceleration 'a'. We know that a collision is to be just avoided, so the distance 'D' should be equal to zero when solving for 'a'.
0.5 * a * t^2 + 44.7 m/s * t - 8.1 m/s * t = 0
Simplifying further, we get:
0.5 * a * t^2 + 36.6 m/s * t = 0
Since 't' cannot be zero (otherwise there would be no collision), we can divide the entire equation by 't':
0.5 * a * t + 36.6 m/s = 0
Simplifying again:
a = - (36.6 m/s) / t
From the previous equations, we can see that t is the time taken for the high-speed train to collide with the locomotive. To avoid a collision, this time must be as small as possible. Therefore, we can assume that t is very close to zero.
So, as t approaches zero, the magnitude of 'a' approaches negative infinity. This means that the magnitude of the required constant deceleration for the high-speed train to avoid a collision with the locomotive is infinitely large.
Therefore, an infinitely large magnitude of deceleration is required to avoid a collision in this scenario.
161 km/h = 44.722 m/s
29 km/hr = 8.056 m/s
Now just solve the usual equations.
The distance traveled by the train must equal the distance traveled by the locomotive, plus the 676m separating them at first:
44.722 + at = 0
44.722t + 1/2 at^2 = 676 + 8.056t
a = -0.946 m/s^2