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