Using the v - t graph, one can conclude: 1. Displacement in (0 - 4) seconds is equal to the displacement in (4 - 6) seconds. 2. Acceleration is 2.5^ -2 ms in OA and zero in AB. 3. Acceleration is negative in time (6 - 8) seconds.

Using the v - t graph, one can conclude: 1. Displacement in (0 - …

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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.

A
Only $A$ is correct.
B
Only $B$ is correct.
✓
Only $C$ is correct.
D
All the three are correct.

✓

Answer

Correct option: C.

Only $C$ is correct.

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 $10\ m$ to $0$. Hence, Statement $C$ is the correct one.

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