The position-vs-time graph shows uniform motion. The object's velocity is

Answer: 1.2\text{m/s} Slope: \Delta x/\Delta t = 12/10 = 1.2\text{m/s}.

The instantaneous velocity at point B on this x-vs-t curve is best found from

Answer: The slope of the <strongtangent line</strong at B. Instantaneous velocity is the slope of the tangent (derivative) at that point. The chord-slope gives <strongaverage</strong velocity over an interval, not instantaneous.

The position-vs-time graph shown represents an object whose acceleration is

Answer: Negative (along −x) The slope (velocity) starts positive and decreases to negative — velocity is decreasing throughout, so acceleration is <strongnegative</strong. The curve bends downward (concave down).

AP Physics 1 · Topic 1.3

Representing Motion Practice

Q: The position-vs-time graph shown represents an object whose acceleration is

Answer: Negative (along −x) The slope (velocity) starts positive and decreases to negative — velocity is decreasing throughout, so acceleration is <strongnegative</strong. The curve bends downward (concave down).

Part of Kinematics.(TOP-1.C)

Practice questions

Want a predicted score for the whole AP AP1 exam? Take the 20-question diagnostic and Lumi will plan the rest.

Take diagnosticarrow_forward Practice all 1.3

Sample questions

5 of 25 — sign in to practice the rest with adaptive difficulty and mastery tracking.

Sample 1difficulty 1/5
$The position-vs-time graph shows uniform motion. The object's velocity is$
- A
  $2.4~\text{m/s}$
- B
  $0.83~\text{m/s}$
- C
  $12~\text{m/s}$
- D
  $1.2~\text{m/s}$
  check_circle
Why
$\Delta x/\Delta t = 12/10 = 1.2~\text{m/s}$
View full solution & statsarrow_forward
Sample 2difficulty 2/5
$x$
- A
 $x$
- B
 $The slope of the tangent line at B.$
 check_circle
- C
 $The area under the curve up to B.$
- D
 $The slope of the chord from origin to B.$
Why
$Instantaneous velocity is the slope of the tangent (derivative) at that point. The chord-slope gives average velocity over an interval, not instantaneous.$
View full solution & statsarrow_forward
Sample 3difficulty 2/5
$The position-vs-time graph shown represents an object whose acceleration is$
- A
 $Negative (along -x)$
 check_circle
- B
 $Cannot be determined$
- C
 $Zero$
- D
 $Positive (along +x)$
Why
$The slope (velocity) starts positive and decreases to negative — velocity is decreasing throughout, so acceleration is negative . The curve bends downward (concave down).$
View full solution & statsarrow_forward
Sample 4difficulty 2/5
$Which position-vs-time graph represents an object moving with constant, positive velocity ?$
- A
 $B$
 check_circle
- B
 $A$
- C
 $None — all are accelerating$
- D
 $C$
Why
$Constant positive velocity gives a straight line with positive, constant slope: graph B. A and C curve, meaning slope (velocity) is changing.$
View full solution & statsarrow_forward
Sample 5difficulty 2/5
$v$
- A
  $t = 8~\text{s}$
- B
  $t = 0~\text{s}$
- C
  $t = 4~\text{s}$
  check_circle
- D
  $It is never at rest$
Why
$t = 4~\text{s}$
View full solution & statsarrow_forward

Predict my AP scorearrow_forward