Car's Final Velocity and Distance Traveled in Braking Period

[Leo34005]

A car starts from rest and travels for 5.9 s with a uniform acceleration of +1.3 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.1 m/s2. The breaks are applied for 1.60 s.
(a) How fast is the car going at the end of the braking period?

(b) How far has the car gone from its start?

 

[G01]

You must make some attempt at the problem to get help on Physics Forums:

https://www.physicsforums.com/showthread.php?t=94379

Where are you getting stuck? What are your thoughts on the problem? What concepts/formulas do you think apply, etc?

 

[tomcue]

I would approach this problem by first identifying the given information and the unknown variables. In this case, we are given the time (5.9 s) and accelerations (+1.3 m/s2 and -2.1 m/s2) and we are asked to find the final velocity and the distance traveled by the car.

To solve for the final velocity (Vf), we can use the equation Vf = Vi + at, where Vi is the initial velocity (which is 0 m/s since the car starts from rest), a is the acceleration, and t is the time. Plugging in the given values, we get Vf = 0 + (1.3 m/s2)(5.9 s) = 7.67 m/s. This means that at the end of the 5.9 seconds of uniform acceleration, the car is traveling at a speed of 7.67 m/s.

To solve for the distance traveled (d), we can use the equation d = Vit + 0.5at2, where Vi is the initial velocity, a is the acceleration, and t is the time. Again, since the car starts from rest, Vi is 0 m/s. Plugging in the given values, we get d = (0)(5.9 s) + 0.5(-2.1 m/s2)(1.60 s)2 = -2.688 m. This means that the car has traveled a distance of 2.688 meters in the opposite direction during the 1.60 seconds of braking.

In summary, at the end of the braking period, the car is traveling at a speed of 7.67 m/s and has traveled a distance of 2.688 meters from its starting point. These calculations are based on the given information and the laws of motion in physics.