Anonymous asked in Science & MathematicsPhysics · 1 decade ago

Suppose that an object is moving with constant acceleration?

Make a statement concerning its motion with respect to time

a. In equal times its velocity changes by equal amounts.

b. In equal times its speed increases by equal amounts.

c. In equal times it moves equal distances.

d. A statement cannot be made using the information given.

5 Answers

  • Andy S
    Lv 6
    1 decade ago
    Favorite Answer


    Because by definition ā = ∆v/∆t

    "change in velocity over change in time"


    ∆v=k*∆t, where k is constant

  • Jim
    Lv 7
    1 decade ago

    As you did not, and the question does not refer to the

    motion as one-dimensional, I'll assume it's at least two-dimensional.

    acceleration is a vector quantity and the changer of velocity

    if acceleration is said to be "constant" it must be so in BOTH

    size AND direction

    in equal time intervals, a constant change is added to the velocity,

    (a) is correct

    since speed is the size of the velocity vector, it too must change equally with time if the acceleration vector is constant.

    (b) is correct also

    b* comment: folks will say the "centripetal acceleration" of circular motion is a "constant accleration" but this only refers to its SIZE, its direction is constantly changing so the VECTOR centripetal acceleration is not constant.

  • 1 decade ago

    constant acceleration means increasing velocity....

    a. In equal times its velocity changes by equal amounts.

  • 1 decade ago

    answer is a.

    it's not b, because in circular motion an object is accelerating even at constant speed.

    it's not c, which implies constant velocity, therefore zero acceleration.

  • How do you think about the answers? You can sign in to vote the answer.
  • Anonymous
    4 years ago

    in equal times its velocity changes by equal amounts As a = ∆v/∆t =>∆v = a x ∆t =>for same ∆t, ∆v will be same as a is constant.

Still have questions? Get your answers by asking now.