It slows down w/an acceleration given by -0.50t in m/s^2 for t in seconds.
lewando said:Could this mean the acceleration is not constant? -0.50t m/s2 is different from -0.50 m/s2.
for problem 2 they want _velocity_ , not distance ...
I don't know . . . you're ignoring units! where did the 6 t come from ? a = 3 m/s^2
The formula for calculating the time when a car stops is t = (v0 - vf) / a, where t is time, v0 is the initial velocity, vf is the final velocity, and a is the acceleration.
The initial velocity of a car can be determined by measuring its speed at the beginning of its motion. This can be done using a speedometer or by measuring the distance the car travels in a certain amount of time.
The greater the acceleration, the shorter the time it will take for a car to stop. This is because acceleration is directly proportional to the change in velocity over time, so a higher acceleration will result in a faster change in velocity and thus a shorter stopping time.
Yes, you can calculate the time when a car stops using the formula t = (vf^2 - v0^2) / (2a), where t is time, vf is the final velocity, v0 is the initial velocity, and a is the acceleration. This formula assumes constant acceleration and can be derived from the original formula by substituting vf = v0 + at.
The weight of a car does not directly affect its stopping time. However, a heavier car will require a greater force to stop, which may result in a longer stopping distance. Additionally, a heavier car may have different acceleration capabilities, which can affect the time it takes for the car to stop.