How to Interpret the Displacement-Time Graphs

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Question 1

(a)     Find the initial displacement.

\( x=0 \)

(b)     Find the displacement at \( t=4\).

\( x=3 \)

(c)     Find the velocity during \( 0 \le t \le 3 \).

\( \displaystyle \text{velocity} = \frac{\text{rise}}{\text{run}} = \frac{3-0}{3-0} = 1 \text{ ms}^{-1} \)

(d)     Find the velocity during \( 3 \le t \le 7 \).

\( \displaystyle \text{velocity} = \frac{\text{rise}}{\text{run}} = \frac{3-3}{7-3} = \frac{0}{4} = 0 \text{ ms}^{-1} \)

(e)     Find the velocity during \( 7 \le t \le 10 \).

\( \displaystyle \text{velocity} = \frac{\text{rise}}{\text{run}} = \frac{0-3}{10-7} = -1 \text{ ms}^{-1} \)

(f)     Find the distance travelled for \( 0 \le t \le 10 \).

\( 3+0+3=6 \text{ m} \)

Question 2

(a)     Find the initial displacement.

\( x=3 \)

(b)     Find the displacement at \( t=4 \).

\( x=-1 \)

(c)     Find the velocity during \( 0 \le t \le 5 \).

\( \displaystyle \text{velocity} = \frac{\text{rise}}{\text{run}} = \frac{-2-3}{5-0} = -1 \text{ ms}^{-1} \)

(d)     Find the time(s) when the velocity becomes zero.

\( 5 \le t \le 8 \)

(e)     Find the time(s) when the velocity becomes positive.

\( 8 \le t \le 10 \)

(f)     Find the time(s) when the particle returns to the origin.

\( t=3 \text{ and } 10 \text{ seconds} \)

(g)     Find the time(s) when the particle moves backward.

\( 8 \le t \le 10 \)

(h)     Find the distance travelled for \( 0 \le t \le 10 \).

\( 5+0+2=7 \text{ m} \)

Question 3

(a)     Find the time(s) when the particle changes its direction.

\( t=2 \text{ and } 6 \text{ seconds} \)

(b)     Find the distance travelled for the first \( 10 \) seconds.

\( 3+6+3=12 \text{ m} \)

(c)     Find the velocity during \( 0 \le t \le 3 \).

\( \displaystyle \text{velocity} = \frac{\text{rise}}{\text{run}} = \frac{3-0}{3-0} = 1 \text{ ms}^{-1} \)

(d)     Find the velocity during \( 3 \le t \le 6 \).

\( \displaystyle \text{velocity} = \frac{\text{rise}}{\text{run}} = \frac{-3-3}{6-3} = -2 \text{ ms}^{-1} \)

(e)     Find the velocity during \( 6 \le t \le 10 \).

\( \displaystyle \text{velocity} = \frac{\text{rise}}{\text{run}} = \frac{3}{4} \text{ ms}^{-1} \)

(f)     Find the maximum speed of the particle.

Speed is the positive value of velocity, thus \( 2 \text{ ms}^{-1} \).

Question 4

(a)     Find the time(s) when the particle changes its direction.

\( t=2 \text{ and } 6 \text{ seconds} \)

(b)     Find the distance travelled for the first \( 10 \) seconds.

\( 4+8+4=16 \text{ m} \)

(c)     Find the time(s) when the particle’s velocity becomes maximum.

velocity is the gradient of \( x \) maximum gradient occurs at \( t=4 \).

(d)     Find the time(s) when the particle moves forward.

\( 2 \lt y \lt 6 \)

Question 5

(a)     Find the speed when the particle reaches the origin.

\( 1 \text{ms}^{-1} \)

(b)     Find the distance travelled for the first \( 10 \) seconds.

\( 10 \text{ m} \)

Question 6

(a)     Find the time(s) when the particle changes its direction.

\( t=2 \text{ and } 8 \text{ seconds} \)

(b)     Find the time(s) when the particle reaches its maximum velocity.

\( t=0 \text{ and } 10 \text{ seconds} \)

 

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