Car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like an arc of a circle of radius 500 m. Now the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant

0.308 m/s2 at an angle of 13.5° below the horizontal

Explanation:

The parallel acceleration to the roadway is the tangential acceleration on the rise.

The normal acceleration is the centripetal acceleration due to the arc. This is given by

The tangential acceleration, from the question, is

The magnitude of the total acceleration is the resultant of the two accelerations. Because these are perpendicular to each other, the resultant is given by

The angle the resultant makes with the horizontal is given by

Note that this angle is measured from the horizontal downwards because the centripetal acceleration is directed towards the centre of the arc

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.