Question

The velocity time graph of two particles moving in a straight line are shown in the adjoining figure Which of the two particles has greater acceleration and how much

Answer 1

acceleration of graph A is greater than that of graph B.

see the diagram, A and B are two graphs drawn between velocity - time co-ordinates.

we know, acceleration is the rate of change of velocity with respect to time.

i.e., a = dv/dt

in velocity-time graph, slope of graph shows acceleration. but we know, slope = tanθ, where θ is angle made by graph with positive x-axis.

in above graph, angle made by A > angle made by B.

⇒slope of graph A > slope of graph B

⇒acceleration of graph A > acceleration of graph B.

hence, acceleration of graph A is greater than that of graph B.

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Answer 2

$\huge\bold\purple{Answer:-}$

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