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NEET 2023 Model Paper 1

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Question : 9 of 200

Marks: +1, -0

The acceleration-time graph for a particle moving along

x

-axis is shown in the figure. The initial velocity of the particle is

2\ \mathrm{m/s}

, then the change in the velocity of particle from

t=0

t=8\ \mathrm{s}

(\mathrm{m/s}^2)

$42\ \mathrm{m/s}$
$38\ \mathrm{m/s}$
$40\ \mathrm{m/s}$
$20\ \mathrm{m/s}$

Solution:

The change in velocity is equal to the area under the acceleration-time graph.

\Delta v = \int_0^8 a(t)\,dt

From the graph, the area is positive from

t=0

to

t=6\ \mathrm{s}

and negative from

t=6\ \mathrm{s}

to

t=8\ \mathrm{s}

.

Positive area:

A_1 = \frac{1}{2}\times 6\times 20 = 60\ \mathrm{m/s}

Negative area:

A_2 = \frac{1}{2}\times 2\times (-20) = -20\ \mathrm{m/s}

Net change in velocity:

\Delta v = A_1 + A_2 = 60 - 20 = 40\ \mathrm{m/s}

The given initial velocity

2\ \mathrm{m/s}

is not needed for finding the change in velocity.

Hence, the correct option is

40\ \mathrm{m/s}

.

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