What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h?
i got the answer to be 13062.5 N. i don't think that's right...
Change velocity from km/h to m/s
95 km/h*(1000 m/km)*(1 h/3600 s)= 26.4 m/s
a=ΔV/Δt=26.4/8=3.3 m/s^2
ΣF=m*a=1100kg*3.3 m/s^2=3630 N
Tada!
First convert kg/h to m/s
95 km/h = (95000 m/h)/(3600) s/h
= 26.4 m/s
Force x Time = momentum change
= 1100*26.4 = 2.90*10^4 kg m/s
Divide that by the time interval (8.0s) to get the average force in Newtons
Well, I hate to say it, but your answer is a bit off. Let me shed some light on the situation and help you calculate the correct force.
To find the force required to stop the car, we need to use Newton's second law: force equals mass times acceleration (F = ma). In this case, the acceleration is the change in velocity divided by the change in time.
First, let's convert the car's initial velocity from km/h to m/s. 95 km/h equals 26.389 m/s (approximately).
Now, we calculate the acceleration. The change in velocity is from 26.389 m/s to 0 m/s, so the change in velocity is -26.389 m/s. The change in time is 8.0 s.
Using the formula a = (Δv) / (Δt), we find that the acceleration is -3.299 m/s² (approximately).
Now, we can use Newton's second law (F = ma) to calculate the force: F = (1100 kg)(-3.299 m/s²) = -3630.9 N (approximately).
So, the average force required to stop the car is approximately -3630.9 N. Keep in mind that the negative sign indicates the direction of the force, which is opposing the car's motion. But hey, look on the bright side – at least the force isn't positive and pushing the car to go faster!
To find the average force required to stop a car, we can use Newton's second law of motion, which states that force is equal to mass times acceleration (F = ma).
First, let's convert the car's speed from km/h to m/s since the SI unit for acceleration is meters per second squared (m/s²).
Given:
Mass of the car (m) = 1100 kg
Initial speed of the car (v) = 95 km/h
Time taken to stop (t) = 8.0 s
To convert km/h to m/s, we can use the conversion factor: 1 km/h = 1000 m/3600 s
Converting the initial speed:
v = 95 km/h * (1000 m/3600 s)
v = 26.39 m/s (approximately)
Now, we can calculate the acceleration using the formula:
Acceleration (a) = (final velocity - initial velocity) / time
Since the car is coming to a stop, the final velocity will be zero.
a = (0 - 26.39 m/s) / 8.0 s
a = -3.299 m/s² (negative sign indicates deceleration)
Now, we can calculate the average force required:
F = m * a
F = 1100 kg * -3.299 m/s²
F = -3628.9 N
The negative sign indicates that the force is in the opposite direction of the car's motion, which is required to stop it.
Therefore, the average force required to stop the car in 8.0 s is approximately 3628.9 N.
Impluse= change in momentum
Force*time= 1100*26.4m/s
I don't get your numbers, about one fourth of that. Did your conversion of km/hr to m/s same as mine?
Here is a neat trick. Put in the google search window...
1100kg*95km/hr* 1/(8seconds) in newtons