Lizi said:Homework Statement
A car moving at 20m/s starts decelerating, travels distance d, and stops. Find the car’s velocity at distance d/2.
Homework Equations
The Attempt at a Solution
my brain is fried. I feel like I’m missing something obvious but I just don’t get it.
neilparker62 said:It was a bit of a brain 'fry' !
Time 'reversing' so that we have acceleration from 0 m/s instead of deceleration from 20 m/s:
$$ d = ½at^2 ⇒ \frac{d}{2} = ½a\left({\frac{t}{\sqrt{2}}}\right)^2 ⇒v=\frac{at}{\sqrt{2}}$$
at = ?stockzahn said:But since you don't know ##a## you can't obtain ##t##, so you stick with two unknowns. I think the question should be solved independent of the time ...
The velocity of a stopping car can be calculated by dividing the initial velocity (speed at the beginning of braking) by the time it takes for the car to come to a complete stop.
The velocity of a stopping car can be affected by factors such as the initial velocity, the mass of the car, the braking force applied, and the coefficient of friction between the tires and the road surface.
Yes, the velocity of a stopping car can be negative if it is moving in the opposite direction of the initial velocity. This is known as deceleration or negative acceleration.
The velocity of a stopping car is directly proportional to its kinetic energy. This means that the higher the velocity, the greater the kinetic energy, and the longer it will take for the car to come to a complete stop.
Velocity and speed are often used interchangeably, but they have different meanings. Velocity is a vector quantity that includes both the speed and direction of an object, while speed is a scalar quantity that only refers to the magnitude of the object's motion. When a car is stopping, its velocity is changing as it slows down, but its speed is decreasing at a constant rate.