Physics, asked by swatisharma82341, 8 months ago

the velocity time graphs of two particles moving in a straight line are shown in the adjoining figure

Answers

Answered by sumitverma17
1

Answer:

s

a

=66.67 m distance traveled by A

s_b=25\ ms

b

=25 m distance traveled by B

Explanation:

1.

Particle A has a greater acceleration because its rate of change is velocity is faster. Looking at line A we observe that it reaches a veliocity of 40 meter per second at the end of 30 seconds.

so, a_a=\frac{40}{30} =\frac{4}{3}\ m.s^{-2}a

a

=

30

40

=

3

4

m.s

−2

2.

After 10 seconds:

velocity of A, v_a=10\ m.s^{-1}v

a

=10 m.s

−1

velocity of B, v_b=4\ m.s^{-1}v

b

=4 m.s

−1

initially the velocity of both the particles is zero:

u_a=u_b=0u

a

=u

b

=0

Acceleration of particle B:

we find that it reaches a speed of 20 meter per second at the end of 40 seconds.

a_b=\frac{20}{40} =0.5\ m.s^{-2}a

b

=

40

20

=0.5 m.s

−2

Now we use the equation of motion:

s_a=u_a.t+0.5\times a_a.t^2s

a

=u

a

.t+0.5×a

a

.t

2

s_a=0+0.5\times 1.33\times 10^2s

a

=0+0.5×1.33×10

2

s_a=66.67\ ms

a

=66.67 m

For B:

s_b=u_b.t+0.5\times a_b.t^2s

b

=u

b

.t+0.5×a

b

.t

2

s_b=0+0.5\times 0.5\times 10^2s

b

=0+0.5×0.5×10

2

s_b=25\ ms

b

=25 m

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