The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. 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) ?
a)
The area under the speed-time graph gives the distance transverse by the particle.
The distance transverse by the particle between t = 0 s to 10 s is given as,
d = 1 2 × 12 × 10 = 60 m
Thus, the distance transverse by the particle between t = 0 s to 10 s is 60 m.
The magnitude of average speed of particle is given as,
v a v = d t
Where, v a v is the average speed of particle, d is the distance transverse by the particle and t is the time taken by the particle.
By substituting the values in the above expression, we get
v a v = 60 m 10 s = 6 ms − 1
Thus, the distance transverse by the particle between t = 0 s to 10 s is 60 m and the average speed of the particle is 6 ms − 1 .
b)
Let s 1 be the distance travelled by the particle from t = 2 s to 5 s and s 2 be the For distance travelled by the particle from t = 2 s to 5 s .
Let u ′ be the velocity of particle after 2 s and a ′ be the acceleration of particle from t = 0 s to 5 s .
The acceleration of the particle by using Newton’s first equation of motion is given as.
v = u ′ + a ′ t
Where, v is the final velocity of the particle, u ′ is the velocity of particle after 2 s , a ′ is the acceleration of particle from t = 0 s to 5 s and t is the time taken by the particle.
By substituting the values in the above expression, we get
12 = 0 + ( a ′ × 5 ) a ′ = 12 5 = 2.4 ms − 2
The final velocity of particle after t = 2 s is given as,
u ′ = u + a ′ t
Where, u ′ is the velocity of particle after 2 s .
By substituting the values in the above expression, we get
u ′ = 0 + ( 2.4 × 2 ) = 4.8 ms − 1
The distance travelled by particle from ( t = ( 5 s − 2 s ) = 3 s ) is given as,
s 1 = u ′ t + 1 2 a ′ t 2
Where, s 1 is the distance travelled by the particle from t = 2 s to 5 s .
By substituting the values in the above expression, we get
s 1 = ( 4.8 × 3 ) + 1 2 ( 2.4 × 3 2 ) = 25.2 m
For distance travelled by the particle from t = 5 s to 6 s .
The acceleration of the particle by using Newton’s first equation of motion is given as.
v = u + a ″ t
Where, v is the final velocity of the particle, u is the velocity of particle after 2 s , a ″ is the acceleration of particle from t = 5 s to 10 s and t is the time taken by the particle.
By substituting the values in the above expression, we get
0 = 12 + ( a ″ × 5 ) a ″ = − 12 5 = − 2.4 ms − 2
The distance travelled by particle from ( t = ( 6 s − 5 s ) = 1 s ) is given as,
s 2 = u t + 1 2 a ″ t 2
Where, s 2 is the distance travelled by the particle from t = 5 s to 6 s .
By substituting the values in the above expression, we get
s 2 = ( 12 × 1 ) + 1 2 ( − 2.4 × 1 2 ) = 10.8 m
The total distance travelled by the particle from t = 2 s to 6 s is,
s = s 1 + s 2
Where, s is the total distance travelled by particle from t = 2 s to 6 s .
By substituting the values in the above expression, we get
s = 25.2 m + 10.8 m = 36 m
The average speed of the particle from t = 2 s to 6 s is given as,
v a v = s t
Where, v a v is the average velocity of the particle.
By substituting the values in the above expression, we get
v a v = 36 m 4 s = 9 ms − 1
Thus, the distance transverse by the particle between t = 2 s to 6 s is 36 m and the average speed of the particle is 9 ms − 1 .
a)
The area under the speed-time graph gives the distance transverse by the particle.
The distance transverse by the particle between
Thus, the distance transverse by the particle between
The magnitude of average speed of particle is given as,
Where,
By substituting the values in the above expression, we get
Thus, the distance transverse by the particle between
b)
Let
Let
The acceleration of the particle by using Newton’s first equation of motion is given as.
Where,
By substituting the values in the above expression, we get
The final velocity of particle after
Where,
By substituting the values in the above expression, we get
The distance travelled by particle from
Where,
By substituting the values in the above expression, we get
For distance travelled by the particle from
The acceleration of the particle by using Newton’s first equation of motion is given as.
Where,
By substituting the values in the above expression, we get
The distance travelled by particle from
Where,
By substituting the values in the above expression, we get
The total distance travelled by the particle from
Where,
By substituting the values in the above expression, we get
The average speed of the particle from
Where,
By substituting the values in the above expression, we get
Thus, the distance transverse by the particle between
