Question
Using the $v- t$ graph, one can conclude:
- Displacement in $(0 - 4)$ seconds is equal to the displacement in $(4 - 6)$ seconds.
- Acceleration is $2.5^{-2}ms$ in $OA$ and zero in $AB.$
- Acceleration is negative in time $(6 - 8)$ seconds.
Using the $v- t$ graph, one can conclude:
✓
Answer
For the statement$(A):$ Displacement in $(0 - 4)s$ increasing and Displacement in $(4 - 6)s$ is Constant.
So, both the displacement of time $(0 - 4)s$ and $(4 - 6)s$ are not equal to each other.
For the statement$(B):$ Acceleration$(OA) =\frac{10}{4}=2.5\text{ms}$
Acceleration(AB) $=\frac{10}{(4+6)}=1\text{ms}$
For the statement$(C):$ As we see that in the graph, part $BC$ is decreasing simultaneously from $10m$ to $0.$
So, both the displacement of time $(0 - 4)s$ and $(4 - 6)s$ are not equal to each other.
For the statement$(B):$ Acceleration$(OA) =\frac{10}{4}=2.5\text{ms}$
Acceleration(AB) $=\frac{10}{(4+6)}=1\text{ms}$
For the statement$(C):$ As we see that in the graph, part $BC$ is decreasing simultaneously from $10m$ to $0.$
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