Consider the motion of the object whose velocity-time graph is given in the diagram below
a) What is the acceleration of the object between times t=0s and t= 2s?
b) What is the acceleration of the object between times t= 10s and t=12 s?
c) What is the net displacement of the object between times t=0s and t=16s.
Answers
Given : velocity-time graph of motion of the object
To find : acceleration of the object between times t=0s and t= 2s , & t= 10s and t=12 s . net displacement of the object between times t=0s and t=16s
Solution:
Time sec Velocity m/s
0 0
2 8
10 8
12 4
16 4
acceleration of the object between times t=0s and t= 2s = ( 8 - 0)/(2 - 0)
= 8/2
= 4 m/s²
acceleration of the object between times t=10s and t= 12s = ( 4 - 8)/(12 - 10)
= -4/2
= -2 m/s²
the net displacement of the object between times t=0s and t=16s.
= (1/2)(0 + 8)(2 - 0) + 8 (10 - 2) + (1/2)(8 + 4)(12 - 10) + 4(16 - 12)
= 8 + 64 + 12 + 16
= 100 m
acceleration of the object between times t=0s and t= 2s = 4 m/s²
acceleration of the object between times t=10s and t= 12s = -2 m/s²
net displacement of the object between times t=0s and t=16s. = 100 m
Learn more:
Figure given below shows the graph of velocity v ofparticle moving ...
https://brainly.in/question/11151179
Figure given below shows the graph of velocity v ofparticle moving ...
https://brainly.in/question/10432386
The graph represents Jason's speed on a recent trip. For which time ...
https://brainly.in/question/15003150
Answer:
Step-by-step explanation:
The $v$-$t$ graph is a straight-line between $t=0$ and $t=2$, indicating constant acceleration during this time period. Hence,
\begin{displaymath}
a = \frac{{\mit\Delta}v}{{\mit\Delta} t} = \frac{v(t=2)-v(t=0)}{2-0} =
\frac{8-0}{2} = 4 {\rm m s^{-2}}.
\end{displaymath}
The $v$-$t$ graph is a straight-line between $t=10$ and $t=12$, indicating constant acceleration during this time period. Hence,
\begin{displaymath}
a = \frac{{\mit\Delta}v}{{\mit\Delta} t} = \frac{v(t=12)-v(t=10)}{12-10} =
\frac{4-8}{2} = -2 {\rm m s^{-2}}.
\end{displaymath}
The negative sign indicates that the object is decelerating.
Now, $v=dx/dt$, so
\begin{displaymath}
x(16)-x(0) = \int_0^{16} v(t) dt.
\end{displaymath}
In other words, the net displacement between times $t=0$ and $t=16$ equals the area under the $v$-$t$ curve, evaluated between these two times. Recalling that the area of a triangle is half its width times its height, the number of grid-squares under the $v$-$t$ curve is 25. The area of each grid-square is $2\times 2=4 {\rm m}$. Hence,
\begin{displaymath}
x(16)-x(0) = 4\times 25 = 100 {\rm m}.
\end{displaymath}