How Do I find acceleration without distance and time?

Question

Detailed question:

Two cars - Car A is going 100 km/h and Car B going 50 km/h

Both have a mass of 1000 kg

How fast is Car A going when Car B stops using breaks? Car A also presses on the breaks --- breaks are pressed at the same time.

Everything is a controlled variable except for the speed.

How do i figure this out using kinematics equations?

All I have is that the car's both have a mass of 1000kg and Car A is travelling at 100 km/h and car B is travelling at 50 km/h

Answer 1

If the two cars are identical in everyway (except for speed) and the break pedal was pushed in both cars at the exact same time and with the same force...both cars would accelerate at the same rate.

If one car is traveling at 50 km/h and comes to a stop in time t, this means that the car experiences some acceleration a during that time t to cause a change in velocity equal to its original speed (delta v = 50 km/h). Since both cars are identical, the second car would also experience the same acceleration, a, during time t, and would therefore experience the same change in speed (delta v = 50 km/h). The only difference here is that the second car will not come to a complete stop after slowing 50 km/h down since its original speed was 100 km/h. Both cars will experience a negative change in their speeds by 50 km/s, but while the first car will stop, the second car will continue to be traveling at 50 km/h (= 100 km/h - 50 km/h) after the ordeal is over.

Answer 2

With the information given in Ur problem the only way to solve it by taking one assumption, that the coefficient of friction between ground and the Tyres of both cars is same. ( this is quite a safe assumption bcos the cars are moving together so they are moving on the same roads. Assume their Tyres r worn out similarly. so the coefficient will be same.)

Now if the friction coefficient is same both cars will decelerate at the same rate. As we know that the decelerations of both cars are same and breaks are applied at the same time then the question is finished !

Since breaks are applied simultaneously then obviously the cars must loss same amount of speed in the same time.So when the car B stops car A is moving at a speed of (100 -50 ) km/hr

But if U do not agree to the assumption then the question will have insufficient information.And U see U do not need the mass of any cars here. Even if they had different masses the answer would be same !( bcos acceleration is always coff. of fric X g )

Answer 3

Mass and velocity.

Answer 4

this sounds like a question from engineering dynamics.