Question

The velocity - time graphs of two particles moving in a straight line are shown in the adjoining figure.
(i) Which of the two particles has greater acceleration and how much?
(ii) Calculate the difference between the distances travelled by the particles in 10 seconds (See Lesson-2)

Answer 1

$s_a=66.67\ m$ distance traveled by A

$s_b=25\ m$ 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, $a_a=\frac{40}{30} =\frac{4}{3}\ m.s^{-2}$

2.

After 10 seconds:

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

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

initially the velocity of both the particles is zero:

$u_a=u_b=0$

Acceleration of particle B:

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

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

Now we use the  equation of motion:

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

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

$s_a=66.67\ m$

For B:

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

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

$s_b=25\ m$

#leanmore

TOPIC: equation of motion

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