A train traveling at a speed of 110 km/h uniformly decelerates for 20 seconds. During that time it travels 450m.
A) What is the acceleration? (m/s^2)
B) What is its speed after deceleration? (km/h)
A ANS = -.805m/s^2
B ANS = 52km
s = 450m
t = 20 seconds
u = 110km/h*1000m/km /60s/min*60 min/h
u = 110000/ 3600
u = 30.55
A) s = ut + 1/2 at^2
a = s - ut/ 1/2 * t^2
a = 450m - 30.55 * 20 sec/ 1/2 * 20^2
a = 450 - 611/ 1/2 * 400
a = -161/ 200
a = -.805m/s^2
I get stuck trying to figure out how to use the formula to get the answer to B which is 52 km.
I believe I would start with the formula: s = v + u/ 2 *t.
I would really appreciate the help with the second part of the question.
Thank you.
v= v₀+at=30.55-0.805•20=14.5 m/s =14.5•3600/100=52.04 km/h
To find the speed of the train after deceleration, you can use the equation:
v = u + a * t
where:
v = final velocity (speed after deceleration)
u = initial velocity (speed before deceleration)
a = acceleration
t = time
Given values:
u = 30.55 m/s
a = -0.805 m/s^2 (negative sign indicates deceleration)
t = 20 seconds
Substituting the values into the equation:
v = 30.55 + (-0.805) * 20
v = 30.55 - 16.1
v = 14.45 m/s
To convert this speed from meters per second to kilometers per hour:
v = (14.45 * 3600) / 1000
v = 52.02 km/h (rounded to two decimal places)
Therefore, the speed of the train after deceleration is approximately 52 km/h.
To find the speed after deceleration, you can use the formula:
v = u + at
Where:
v = final velocity (unknown)
u = initial velocity (110 km/h)
a = acceleration (-0.805 m/s^2, which we convert to km/h^2)
t = time (20 seconds)
First, let's convert the acceleration from m/s^2 to km/h^2:
1 m/s^2 = 3.6 km/h^2
So, -0.805 m/s^2 * 3.6 km/h^2/m/s^2 = -2.898 km/h^2
Now, let's substitute the values into the formula:
v = 30.55 km/h + (-2.898 km/h^2 * 20 s)
v = 30.55 km/h - 57.96 km/h
v = -27.41 km/h
The negative sign indicates that the train is moving in the opposite direction, as it is decelerating. To get it as a positive value, we can take the absolute value:
v = |-27.41 km/h|
v = 27.41 km/h
So, the speed after deceleration is 27.41 km/h.
Apologies for the misunderstanding in my previous response. The correct answer is 27.41 km/h, not 52 km/h.