Draw the $x\to t$ graphs for positive, negative and zero acceleration.
Draw the $x\to t$ graphs for positive, negative and zero acceleration.
The average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is $v_{0}$ at $t=0$ and $v$ at time $t$,
$\bar{a}=\frac{v-v_{0}}{t}$
Similar Questions
A particle moves in a straight line and its position $x$ at time $t$ is given by $x^2=2+t$. Its acceleration is given by
A particle moves in a straight line and its position $x$ at time $t$ is given by $x^2=2+t$. Its acceleration is given by
The displacement $(x)$ - time $(t)$ graph of a particle is shown in figure. Which of the following is correct?
The displacement $(x)$ - time $(t)$ graph of a particle is shown in figure. Which of the following is correct?
Velocity of a particle is in negative direction with constant acceleration in positive direction. Then, match the following columns.
Colum $I$
Colum $II$
$(A)$ Velocity-time graph
$(p)$ Slope $\rightarrow$ negative
$(B)$ Acceleration-time graph
$(q)$ Slope $\rightarrow$ positive
$(C)$ Displacement-time graph
$(r)$ Slope $\rightarrow$ zero
$(s)$ $\mid$ Slope $\mid \rightarrow$ increasing
$(t)$ $\mid$ Slope $\mid$ $\rightarrow$ decreasing
$(u)$ |Slope| $\rightarrow$ constant
Velocity of a particle is in negative direction with constant acceleration in positive direction. Then, match the following columns.
| Colum $I$ | Colum $II$ |
| $(A)$ Velocity-time graph | $(p)$ Slope $\rightarrow$ negative |
| $(B)$ Acceleration-time graph | $(q)$ Slope $\rightarrow$ positive |
| $(C)$ Displacement-time graph | $(r)$ Slope $\rightarrow$ zero |
| $(s)$ $\mid$ Slope $\mid \rightarrow$ increasing | |
| $(t)$ $\mid$ Slope $\mid$ $\rightarrow$ decreasing | |
| $(u)$ |Slope| $\rightarrow$ constant |
A Body moves $6\, m$ north. $8 \,m$ east and $10\;m$ vertically upwards, what is its resultant displacement from initial position
A Body moves $6\, m$ north. $8 \,m$ east and $10\;m$ vertically upwards, what is its resultant displacement from initial position
A particle of unit mass undergoes one dimensional motion such that its velocity varies according to $ v(x)= \beta {x^{ - 2n}}$, where $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$, is given by
A particle of unit mass undergoes one dimensional motion such that its velocity varies according to $ v(x)= \beta {x^{ - 2n}}$, where $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$, is given by
- [AIPMT 2015]