The position of a particle moving along x-axis is given by x=3t2−t3; where x is in m and t is in sec. Consider the following statements:
The position of a particle moving along x-axis is given by x=3t2−t3; where x is in m and t is in sec. Consider the following statements:
The correct options are
B Distance travelled by the particle upto 4s is 24 m
C Displacement of the particle after 4s is (-16m)
Give: x=3t2−t3
where x is displacement. We need to find distance. We know if the particle is doing 1D motion and its velocity is in one particular direction then its distance and displacement are same. But the moment the particle reverses its direction of motion the displacement starts to decrease but path length or the distance still increases. So we need to find the point where the particle changes its direction of motion or its velocity changes sign.
x=3t2−t3
v=6t−3t2(differentiating)
v=3t(2−t)
v=0 at t=0 and at t=2
The v-t graph will look like the above. I am sure you know how to dram the graph of a quadratic equation. Here a = -ve and hence the inverted graph.
So velocity is +ve from 0 - 2.
So displacement = distance
So in this time interval displacement = distance = [3(2)2−(2)3]−[3(0)2−(0)3]=4
Now from 2 to 4 velocity is reducing
So displacement will reduce.
But distance will increase.
So from 2 to 4 distance will be -ve of displacement.
Distance = -(displacement) = −[3t2−t3]
= −[3(4)2−(4)3]−[3(2)2−(2)3]
Distance = 20
So total distance from t = 0 to t = 4 is 4 + 20 = 24m
Alternate solution:
x=3t2−t3
∴ Displacement at t = 4s, x = -16 m
To find distance
dxdt=v=6t−3t2
When v=0,6t−3t2=0
⇒3t(2−t)=0
⇒t=2,0
.Speed time graph of this motion will be like:
So the graph is same in the time interval 0-2 sec it's the mirror image on the x - axis
⇒ Speed=6t−3t2[fort=(0−2)sec]
= −(6t−3t2) [ for t = (2 - 4) sec ]
Distance = 2∫0speeddt+4∫2speeddt
∴distance=2∫0(6t−3t2)dt−4∫2(6t−3t2)dt
= (6t22−3t33)20−(6t22−3t33)42
= (3t2−t3)20−(3t2−t3)42
= ((3×4)−8)−[(3(6)−64)−(3(4)−8))]
= (12 - 8) - [48 - 64 - ( 12 - 8)]
= 4 - [ - 16 - 4 ]
= 4 + 20
= 24 m.
B
Distance travelled by the particle upto 4s is 24 m
C
Displacement of the particle after 4s is (-16m)
Give: x=3t2−t3
where x is displacement. We need to find distance. We know if the particle is doing 1D motion and its velocity is in one particular direction then its distance and displacement are same. But the moment the particle reverses its direction of motion the displacement starts to decrease but path length or the distance still increases. So we need to find the point where the particle changes its direction of motion or its velocity changes sign.
x=3t2−t3
v=6t−3t2(differentiating)
v=3t(2−t)
v=0 at t=0 and at t=2
The v-t graph will look like the above. I am sure you know how to dram the graph of a quadratic equation. Here a = -ve and hence the inverted graph.
So velocity is +ve from 0 - 2.
So displacement = distance
So in this time interval displacement = distance = [3(2)2−(2)3]−[3(0)2−(0)3]=4
Now from 2 to 4 velocity is reducing
So displacement will reduce.
But distance will increase.
So from 2 to 4 distance will be -ve of displacement.
Distance = -(displacement) = −[3t2−t3]
= −[3(4)2−(4)3]−[3(2)2−(2)3]
Distance = 20
So total distance from t = 0 to t = 4 is 4 + 20 = 24m
Alternate solution:
x=3t2−t3
∴ Displacement at t = 4s, x = -16 m
To find distance
dxdt=v=6t−3t2
When v=0,6t−3t2=0
⇒3t(2−t)=0
⇒t=2,0
.Speed time graph of this motion will be like:
So the graph is same in the time interval 0-2 sec it's the mirror image on the x - axis
⇒ Speed=6t−3t2[fort=(0−2)sec]
= −(6t−3t2) [ for t = (2 - 4) sec ]
Distance = 2∫0speeddt+4∫2speeddt
∴distance=2∫0(6t−3t2)dt−4∫2(6t−3t2)dt
= (6t22−3t33)20−(6t22−3t33)42
= (3t2−t3)20−(3t2−t3)42
= ((3×4)−8)−[(3(6)−64)−(3(4)−8))]
= (12 - 8) - [48 - 64 - ( 12 - 8)]
= 4 - [ - 16 - 4 ]
= 4 + 20
= 24 m.
