The speed of a goods truck which has been shunted on to a level siding falls from 10km/h to 5km/h in moving a distance of 30m. If the retardation is constant, how much further will the truck travel before coming to to rest?
10 km/h * 1000/3600 = 2.78 m/s
5 km/h = 1.39 m/s
v = V i + a t
1.39 = 2.78 + a t
so
a = -1.39/t
x = Xi + Vi t + (1/2) a t^2
30 = 0 + 2.78 t - 0.694 t = 2.09 t
t = 14.4 seconds
a = -1.39/14.4 = - 0.0965 m/s^2
x = average speed during stop * time to stop
average speed = 1.39/2= 0.695 m/s
0 = 1.39 - 0.0965t
t = 14.4 s
x = 0.695 * 14.4
= 10 meters
To find out how much further the truck will travel before coming to rest, we need to calculate the stopping distance. Here's how you can do it step by step:
Step 1: Calculate the initial velocity of the truck (u) in m/s.
Given that the initial velocity of the truck is 10 km/h, we need to convert it to m/s:
10 km/h × (1000 m/1 km) × (1 h/3600 s) = 2.78 m/s
Step 2: Calculate the final velocity of the truck (v) in m/s.
Given that the final velocity of the truck is 5 km/h, we need to convert it to m/s:
5 km/h × (1000 m/1 km) × (1 h/3600 s) = 1.39 m/s
Step 3: Calculate the acceleration (a) of the truck in m/s².
The truck experiences a negative acceleration since it is slowing down.
We can use the equation:
v² = u² + 2as
Where:
v = final velocity = 0 m/s (since the truck comes to rest)
u = initial velocity = 2.78 m/s
a = acceleration
s = distance moved = 30 m
Rearranging the equation, we have:
a = (v² - u²) / (2s)
a = (0 - (2.78)²) / (2 * 30)
a = -2.78² / 60
a ≈ -0.128383 m/s²
Step 4: Calculate the stopping distance (s) of the truck in meters.
We can use the equation:
v² = u² + 2as
Where:
v = final velocity = 0 m/s (since the truck comes to rest)
u = initial velocity = 2.78 m/s
a = acceleration ≈ -0.128383 m/s²
s = stopping distance (to be calculated)
Rearranging the equation, we have:
s = (v² - u²) / (2a)
s = (0 - (2.78)²) / (2 * (-0.128383))
s ≈ -2.78² / -0.256766
s ≈ 34.05 m (rounded to two decimal places)
Therefore, the truck will travel approximately 34.05 meters further before coming to rest.
To determine how much further the truck will travel before coming to rest, we need to use the equation of motion.
The equation of motion for uniformly accelerated linear motion can be expressed as:
(vf^2 - vi^2) = 2ad
Where:
- vf is the final velocity,
- vi is the initial velocity,
- a is the acceleration, and
- d is the distance covered.
In this case, the initial velocity (vi) is 10 km/h, or 10,000 m/3600 s ≈ 2.78 m/s.
The final velocity (vf) is 5 km/h, or 5,000 m/3600 s ≈ 1.39 m/s.
The distance covered (d) is 30 m.
Since the truck is decelerating (slowing down), the acceleration (a) will be negative. The equation becomes:
(1.39^2 - 2.78^2) = 2a(30)
Simplifying the equation gives:
(1.93 - 7.73) = -60a
-5.8 = -60a
To solve for the acceleration (a), divide both sides of the equation by -60:
a ≈ 0.097 m/s^2
Now, we can find the distance (d') traveled from the moment the truck's speed becomes zero until it comes to a complete stop:
vf^2 = vi^2 + 2ad'
Since the final velocity (vf) is zero, the equation becomes:
0 = 2(0.097)d'
Solving for d', we get:
d' = 0
Therefore, the truck will not travel any further before coming to rest.
In conclusion, the truck will travel an additional 30 meters before coming to rest.