phy position and movement


1. The above graph shows how the square of velocity of an object undergoes uniform acceleration varies with

displacement. The object is initially at rest and travels along straight line.

The acceleration of the object is



Y has a linear relationship with time as shown. Y may represent

(1) the speed of a body starting from rest under a constant force.

(2) the distance travelled by a body at constant speed.

(3) the acceleration of a body falling from rest.

3.1st statement 2nd statement

An object thrown vertically upwards has zero The object is instantaneously at rest at its highest point.

acceleration at its highest point.

4.Which of the following is/are vectors?

(1) momentum

(2) kinetic energy

(3) potential difference



The graph above shoes the strobe photograph of a ball rolling down a slope. The stroboscope is flashing at a

frequency of 5 Hz. Find the acceleration of the ball.

Please show steps clearly


1. 1ms-2

2. (1),(2)

3.both false



1 個解答

  • 天同
    Lv 7
    1 十年前

    1. The line on the graph follows the equation of motion: v^2 = u^2 + 2a.s with u = 0 m/s.

    Hence, slope of the line = 2a

    Since from the graph, v^2 = 4 (m/s)^2 when s = 2 m

    Thus, a = (1/2) x slope of line = (1/2) x 4/2 m/s2 = 1 m/s2

    2. Statement (1): v = at, force constant imples a is also constant.Statement (2): s = v.t, if v is constant, s varies linearly with time tStatment (3) is wrong. The acceleration due to gravity is constant independent of time .

    3. First statment is false. The acceleration at any time and position is equal to the acceleration due to gravity, which is not zero.Second statment is right. The object has zero velocity at the highest point.

    4. Momentum has both magnitude and direction, hence is a vector.Kinetic energy and potnetial difference have magnitudes only. They don't have a direction, hence are scalars.

    5. Time between successive images = 1/5 s = 0.2 s

    Consider the first and second images,use equation of motion: s = ut + (1/2)at^2

    4 = (0.2)u + (1/2)a.(0.2)^2 --------------- (1)

    Consider the first and third images, use the equation s = ut + (1/2)at^2

    (4+8) = (0.2+0.2)u + (1/2)a(0.2+0.2)^2

    i.e. 12 = 0.4u + (1/2)a(0.4)^2 ---------------------- (2)

    (2) - (1)x2: 4 = 0.08a - 0.04a

    hence, a = 4/0.04 cm/s2 = 100 cm/s2 = 1 m/s2