1. A train traveling at a speed of 110km/h decelerates for 20 seconds. During that time it travels 450 meters.
A) What is the acceleration? = m/s^2
B) What is its speed after deceleration? = km/h
Initial velocity (u) = 110km/h
Time (t) = 20s
Distance = 450m
Using this formula s = ut + 1/2 at^2
I need some help. Thank you.
v₀=110 km/h=30.6 m/s
s=v₀t-at²/2,
a=2(v₀t-s)/t²=…
v= v₀-at=…
5. An observer measured an interval of 10s between seeing a lightning flash and hearing the thunder. If the temperature of the air was 20 degrees Celsius, how far away was the source of sound?
To find the acceleration (a), you can use the formula:
s = ut + 1/2 at^2
Given:
u = initial velocity = 110 km/h
t = time = 20 seconds
s = distance traveled = 450 meters
Since the initial velocity (u) is given in km/h, we need to convert it to meters per second (m/s) to be consistent with the other units. So, we can convert the initial velocity (u) from km/h to m/s by multiplying it by 1000/3600, which is (1000 meters) / (3600 seconds).
u = 110 km/h * (1000 meters / 3600 seconds) = 30.56 m/s
Now, we can substitute the given values into the formula:
450m = (30.56 m/s)(20s) + (1/2)a(20s)^2
We can simplify this equation by calculating the values on the right side:
20s * (1/2)a(20s) = (1/2) * a * (20^2) * s = 200a
So, the equation becomes:
450m = 30.56 m/s * 20s + 200a
Now, we can rearrange the equation to solve for a:
450m - 30.56 m/s * 20s = 200a
450m - 611.2 m = 200a
-161.2m = 200a
a = (-161.2m) / 200
Now, we can calculate the acceleration:
a = -0.806 m/s^2 (rounded to three decimal places)
To find the speed after deceleration, we need to subtract the acceleration (a) from the initial velocity (u). But before that, we should convert u from m/s back to km/h:
u = 30.56 m/s * (3600 seconds / 1000 meters) = 110 km/h (approximately)
Therefore, the speed after deceleration is:
Speed after deceleration = u - a = 110 km/h - 0.806 m/s^2
The speed after deceleration is approximately 109.194 km/h (rounded to three decimal places).