1. The speed – time graph of a particle moving along a fixed direction is shown
in the fig. Obtain the distance traversed by the particle between (a) t = 0 s
to 10 s, (b) t = 2 s to 6 s. What is the average speed of the particle over the
intervals in (a) and (b)?
[Ans. : (a) 60 m, 6 m/s, (b) 24 m, 6 m/s.]
Speed (m/s)
10
Time (s)
Answers
Answer:
Ki9
Explanation:
Hey mate,
# Answer-
(a) From t=0 to t=10s, s=60m, v=6m/s
(a) From t=2s to t=6s, s=36m, v=9m/s
# Explaination-
(a) Distance travelled by the particle = Area under the given graph
= (1/2)×(10–0)×(12–0) = 60 m
Average speed = Distance / Time = 60/10 = 6 m/s
(b) Let s1 and s2 be the distances covered by the particle between time
t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s
s = s1 + s2 … (i)
For distance s1:
Let u′ be the velocity of the particle after 2 s and a′ be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:
v = u + at
Where,
v = Final velocity of the particle
12 = 0 + a′×5
a′ = 12/5 = 2.4 ms-2
Again, from first equation of motion, we have
v = u + at
= 0 + 2.4×2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
s1 = u‘t + (1/2)a‘t^2
= 4.8 × 3 + (1/2)×2.4×(3)^2
= 25.2 m ……..(ii)
For distance s2:
Let a″ be the acceleration of the particle between time t = 5 s and t = 10 s.
From first equation of motion,
v = u + at (where v = 0 as the particle finally comes to rest)
0 = 12 + a″×5
a″ = -12/5 = – 2.4 ms-2
Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)
s2 = u“t + (1/2)a″t^2
= 12 × 1 + (1/2)(-2.4)×(1)^2
= 12 – 1.2 = 10.8 m ………(iii)
From equations (i), (ii), and (iii), we get
s = 25.2 + 10.8 = 36 m
Average speed = Distance / time = 36 / 4 = 9 m/s
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Answer:
( a ) Distance covered by the particle = Area of the given graph
= (1/2)base x height
= (1/2) x (10) x (12) = 60m
Average speed of the particle = 60/10 = 6 m/s
( b ) The distance traversed by the particle between
t = 1s to 8s
let distance travelled in 1 to 5s be S1 and distance travelled from 6 to 8s be S2.
Thus, total distance travelled, S ( in t = 1 to 8 s) = S1 + S2 . . . . . . ( 1 )
Now, For S1.
Let u’ be the velocity of the particle after 1 second and a’ be the acceleration in the particle from t = 0 to 5s
We know that the particle is under uniform acceleration from t = 0 to 5s thus, we can obtain acceleration using the first equation of motion.
v = u + at
where, v = final velocity
12 = 0 + a’(5)
a’ = 2.4 m/s2
Now to find the velocity of the particle at 1s
v = 0 + 2.4 (1)
v = 2.4 m/s = u’ at t = 1s
Thus, the distance covered by the particle in 4 seconds i.e., from t = 1 to 5 s.
S1 = u’t + ½ a’t2
= 2.4 x 4 + ½ x 2.4 x 42
= 9.6 + 19.2 =28.8 m
Now, for S2
Let a’’ be the uniform acceleration in the particle from 5s to 10s
Using the first law of motion
v = u + at . . . . . . . ( v= 0 as the particle comes to rest )
0 = 12 + a’’ x 5
a’’ = -2.4 m/s
Thus, distance travelled by the particle in 3 seconds i.e., between 5s to 8s
S2 = u’’t + ½ a’’t
S2 = 12 x 3 + ½ x(-2.4) x 32
= 36 + (-1.2)x9
S2 = 25.2m
Thus, putting the values of S1 and S2 in equation ( 1 ), we get:
S = 28.8 + 25.2 = 54m
Therefore, average speed = 54 / 7 = 7.71 m/s.
Explanation: