What braking force is needed to stop a 1000 kilogram car moving at 10 m/s in a time of 1 seconds?

Answer: The average braking force is 2000 N. Answer: The average braking force is 2000 N.

What is the average braking force of a 1000 kg car moving at 10 m/s braking to a stop in 5 s?

The average braking force of a 1000-kg car moving at 10 m/s braking to a stop in 5 s is 2000 N.

What total distance would you travel from when you first apply the brakes until the car stops?

Braking distance is the distance it takes to stop your vehicle once you apply the brakes. At 65 mph, it takes an additional 5.5 seconds or about 525 feet of actual brake application to stop your vehicle.

How far does the car move during the braking?

Driver Care – Know Your Stopping Distance

Speed	Perception/Reaction Distance	Braking Distance
30 mph	44 feet	45 feet
40 mph	59 feet	80 feet
50 mph	73 feet	125 feet
60 mph	88 feet	180 feet

What is the braking force of a car?

Braking Force Definition Braking force is defined as the total force required to stop a car at a set stopping distance when the car is traveling at a known constant velocity.

How do you calculate the braking distance of a car?

Formula for calculating the braking distance. The following formula has proven to be useful for calculating the braking distance: (Speed ÷ 10) × (Speed ÷ 10). At a speed of 100 km/h the braking distance is therefore a full 100 metres..

What is the speed of a car with a mass of 1000kg?

A car whose mass is 1000 kg is traveling at a constant speed of 10 m/s. Neglecting any friction, how much force will the engine have to supply to keep going the same speed? F=ma – Quora A car whose mass is 1000 kg is traveling at a constant speed of 10 m/s.

What is the friction force of 75N over 200 meters?

So your problem basically tells us that the friction force is 75N. 75N applied over 200 meters is 15 kJ. That seems awfully high to me, but it’s where the problem leads.

How do you calculate the change in velocity and acceleration?

The change in velocity is v (final) – v (initial) which is 50–10 = 40 m/s (FORWARD). The acceleration a = (delta v / delta t) is this change in velocity change divided by the time interval, or 2 m/s^2 (FORWARD). the NET FORCE required to do this is F (net) = ma = 1000 * 2 = 2000 N [FORWARD].

How does friction affect the speed of a bike?

The only way you can have constant speed while applying a force is if the friction and other losses are completely absorbing the work done by that force. If losses were lower than that, the bike would accelerate. So your problem basically tells us that the friction force is 75N. 75N applied over 200 meters is 15 kJ.