What average force is required to stop an 1300 -kg car in 9.0 s if the car is traveling at 90 km/h?
Vf=Vi+a*t where a= Force/mass
change km/hr to m/s
What average force is required to stop an 1300 kg car in 9.0 s if the car is traveling at 65 km/h?
To find the average force required to stop the car, you need to use Newton's second law of motion, which states that force equals mass multiplied by acceleration (F = m * a). In this case, the car's mass (m) is given as 1300 kg.
First, you need to find the car's final velocity, which will be 0 since it is coming to a stop. To do this, you need to convert the initial velocity from kilometers per hour to meters per second, as the SI unit for velocity is meters per second.
Given:
- Initial velocity (u) = 90 km/h
- Time (t) = 9.0 s
1. Convert the initial velocity from km/h to m/s:
To convert from km/h to m/s, divide by 3.6 because 1 km/h is equal to 1/3.6 m/s.
So, u = (90 km/h) / (3.6 m/s) = 25 m/s (rounded to 2 decimal places)
2. Calculate the acceleration (a):
Use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Rearrange the equation to solve for acceleration:
a = (v - u) / t
Since v = 0 m/s (as the car comes to a stop), the equation becomes:
a = (0 m/s - 25 m/s) / 9.0 s
a = -25 m/s / 9.0 s (note the negative sign indicates that the acceleration is opposite to the initial velocity)
3. Calculate the average force (F):
Use Newton's second law of motion, F = m * a:
F = (1300 kg) * (-25 m/s / 9.0 s) = -3611 N
So, the average force required to stop the 1300 kg car in 9.0 s is approximately 3611 N.