AP Physics B & C - Multiple Choice Practice Questions on One Dimensional Kinematics

"Whenever you are confronted with an opponent, conquer him with love."

– Mahatma Gandhi

Let us discuss a few interesting multiple choice practice questions on one dimensional motion. Here are some questions beneficial for AP Physics B as well as AP Physics C aspirants:

(1) The adjoining figure shows the velocity time graph of an object. Total displacement suffered by the object during the interval when it has non-zero acceleration and retardation is

(a) 80 m

(b) 70 m

(c) 60 m

(d) 40 m

(e) 30 m

The object has non-zero acceleration and retardation during the time intervals from 5 sec to 15 sec and from 20 sec to 30 sec. The total area under the velocity time graph during these intervals gives the required displacement.

Displacement from 5 sec to 15 sec = 30 m

Displacement from 20 sec to 30 sec = 40 m

Therefore, total displacement = 70 m

(2) Successive positions (x) of an object (moving from left to right) at equal time intervals are shown in the following figure:

Which one among the following position-time graphs best represents the motion of the object? 

 

Change of position is slowest in the beginning and in the end. The motion is best represented by graph (c).

[Note that graph (d) is not the answer since the change of position in the beginning and in the end is shown as fastest in it]

(3) Which one among the following velocity-time graphs best represents the motion of the object mentioned in question no.(2)? 

The graph (c) is the answer.

The following questions are meant for AP Physics C aspirants:

(4) A particle projected vertically upwards attains the maximum height h in time t. While returning from the highest point it takes an additional time t₁ to fall to the height h.2. If air resistance is negligible, how is t₁ related to t?

(a) t₁ = t/2

(b) t₁ = √(t/2)

(c) t₁ = t/√2

(d) t₁ = t/√3

(e) t₁ = t/3

For the upward motion we have

             0 – u² = – 2gh …………(i)

[We have used the equation of motion, v² – u² = 2as. The sign of the gravitational acceleration g is negative since it is opposite to the direction of the velocity of projection u]

Using the equation of motion, v = u + at we have

             0 = u – gt from which u = gt

Substituting this value of u in Eq (i), we have

             g²t² = 2gh

Therefore, t =√(2h/g) ……..(ii)

For the fall through h/2 from the highest point we have

             h/2 = 0×t₁ + ½ gt₁²

[We have used the equation of motion, s = ut + ½ gt²]

Therefore, t₁ =√(h/g)

Comparing this with the value of t given in Eq (ii) we obtain t₁ = t/√2.

(5) A ball projected vertically up has the same vertical displacement h at times t₁ second and t₂ second. If air resistance is negligible, the maximum height reached by the ball is

(a) gt₁t₂/2

(b) g(t₁² + t₂²)/2

(c) g(t₁ + t₂)²/4

(d) g(t₁ + t₂)²/8

(e) 2g(t₁ + t₂)²

The time taken by the ball to move from height h to the top of its trajectory and back to the height h is t₂ – t₁. Therefore, the time taken to move from height h to the top of the trajectory is (t₂ – t₁)/2.

The total time taken for the upward journey (from ground to the top most point) is evidently t₁ + t₂ – t₁)/2 = (t₁ + t₂)/2.

Time taken for the return journey (from the top most point to the ground) also is equal to (t₁ + t₂)/2. Therefore, the maximum height H (using the equation, s = ut + ½ gt²) is given by

             H = 0 + ½ g [(t₁ + t₂)/2]²

Or, H = g(t₁ + t₂)²/8

             Questions on kinematics were discussed earlier on this site. You can access them either by clicking on the label ‘kinematics’ below this post or by trying a search for                  ‘kinematics’ using the search box provided on this page.