What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h?
i understand what you both did, but why can't you use the V = Vo + at equation and solve for acceleration there and use that in the EF = ma equation?
Using the equation V = Vo + at to solve for acceleration (a) and then substituting it into the equation F = ma may seem like a valid approach at first. However, in this specific scenario, this method would not provide an accurate solution.
Let me explain why:
To find the average force required to stop the car, we need to calculate the deceleration (negative acceleration) first. In this case, the car is slowing down, so the acceleration is in the opposite direction to its initial velocity Vo.
Using the equation V = Vo + at:
V = final velocity = 0 m/s (the car comes to a stop)
Vo = initial velocity = 95 km/h = 26.4 m/s (converted from km/h to m/s)
t = time taken to stop = 8.0 s
Plugging the values into the equation, we can solve for acceleration:
0 = 26.4 + a * 8.0
a = (0 - 26.4) / 8.0
a = -3.3 m/s^2
Now, according to Newton's second law of motion, the force (F) required to stop the car can be calculated using the equation F = ma, where m is the mass of the car.
Given:
m = 1100 kg
a = -3.3 m/s^2
F = (1100 kg) * (-3.3 m/s^2)
F = -3630 N
The negative sign signifies that the force required acts in the opposite direction to the motion of the car. This means that to stop the car, a force of 3630 Newtons, directed opposite to its motion, is needed.
In summary, using the equation V = Vo + at to calculate acceleration and substituting it into the equation F = ma is not appropriate in this case because the car's final velocity (0 m/s) is already known. The correct approach is to directly calculate the deceleration using the time taken to stop and then use Newton's second law to find the force required.
Sure. a= averageforce/m
Then solve for average force. Same equation.