2. The position (x)-time (t) graph for a particle moving
along a straight line is shown in figure. The average
speed of particle in time interval t = 0 to t = 8 s is
Answers
Answer:
speed = d/t
d =??
area of triangle 0to 2 sec =10
ar of rectangle from 2 to 4 sec=20,area from 4 to 6 sec=30
ar from 6 to 8 sec=20
total area = 80
that is total distance
speed =80/8 = 10 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: