Question

the v-t graph of a linear motion is shown in adjoining figure. the distance from origin after 8 seconds is

Answer 1

Distance =area under the graph...

So.
Distance=[(4×1)/2]+(2×4)+[(4×1)/2]+[(-2×1)/2]+[(-2)×2]+[(-2×1)/2]
=2+8+2-1+-4+-1
=12-6
=6m

Hence the correct answer is D. 6meters.

Hope this is what you were looking for...

Regards,
Leukonov.

Answer 2

Answer:

The distance from the origin after 8 seconds is 6 meters.

Explanation:

The area under the velocity-time graph(v-t) gives the distance.

Distance(d)= Area under v-t graph

CASE 1 : When t= 1 second

The velocity(v)=4 m/s

The time(t) =1 second

$\mathrm{d_1=\frac{1}{2} \times v\times t}$                (1)

Inserting the value of "v" and "t" in equation (1) we get;

$\mathrm{d_1=\frac{1}{2} \times 4\times 1}$

$\mathrm{d_1=2\ meters}$         (2)

CASE 2 : When t= 1 to 3 second

The velocity(v)=4 m/s

The time(t) =3 second -1 second= 2 second

$\mathrm{d_2=v\times t}$        (3)

Inserting the value of "v" and "t" in equation (3) we get;

$\mathrm{d_2=4\times 2}$

$\mathrm{d_2=8\ meters}$   (4)

CASE 3: When t= 3 to 4 second

The velocity(v)=4 m/s

The time(t) =4 second -3 second= 1 second

$\mathrm{d_3=\frac{1}{2} \times v\times t}$                (5)

Inserting the value of "v" and "t" in equation (5) we get;

$\mathrm{d_3=\frac{1}{2} \times 4\times 1}$

$\mathrm{d_3=2\ meters}$           (6)

CASE 4: When t= 4 to 5 second

The velocity(v)=-2 m/s

The time(t) =5 second - 4second= 1 second

$\mathrm{d_4=\frac{1}{2} \times v\times t}$          (7)

Inserting the value of "v" and "t" in equation (7) we get;

$\mathrm{d_4=\frac{1}{2} \times -2\times 1}$

$\mathrm{d_4=-1\ meters}$       (8)

CASE 5: When t= 5 to 7 second

The velocity(v)=-2 m/s

The time(t) =7 second - 5 second= 2 second

$\mathrm{d_5=v\times t}$        (9)

Inserting the value of "v" and "t" in equation (7) we get;

$\mathrm{d_5=-2\times 2}$

$\mathrm{d_5=-4\ meters}$       (10)

CASE 6: When t= 7 to 8 second

The velocity(v)=-2 m/s

The time(t) =8 second - 7 second= 1 second

$\mathrm{d_6=\frac{1}{2} \times v\times t}$          (11)

Inserting the value of "v" and "t" in equation (11) we get;

$\mathrm{d_6=\frac{1}{2} \times -2\times 1}$

$\mathrm{d_6=-1\ meters}$      (12)

We will get the total distance after adding equations (2),(4),(6),(8),(10), and (12);

$\mathrm{d=d_1+d_2+d_3+d_4+d_5+d_6}$         (13)

By inserting the value of "d₁","d₂","d₃","d₄","d₅", and "d₆" we get;

$\mathrm{d=2+8+2-1-4-1}$

$\mathrm{d=12-6}$

$\mathrm{d=6\ meters}$        (14)

So, the distance from the origin after 8 seconds is 6 meters.

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