The position x of particle varies with time t as x = 6 + 12t – 2t² where x is in metre and t in second. What is the distance travelled by the particle in first 5 seconds ?
Answers
Answer :-
Distance = 26 m
Explanation :-
Position of the particle is given by -
x = 6 + 12t – 2t²
Differentiating w . r . t . t,
dx / dt = 12 - 2 × 2t
v = 12 - 4t
When particle stops travelling in forward direction, v = 0 :
0 = 12 - 4t
t = 12 / 4
t = 3 s
Position of the particle at t = 0 s,
x = 6 + 12t – 2t²
x ^ 0 = 6 + 12 × 0 - 2 × 0²
x ^ 0 = 6 m
Position of the particle at t = 3 s,
x = 6 + 12t' – 2t'²
x³ = 6 + 12 × 3 – 2 × 3²
x³ = 6 + 36 - 2 × 9
x³ = 24 m
Position of the particle at t = 5 s,
x = 6 + 12t' – 2t'²
x ^ 5 = 6 + 12 × 5 – 2 × 5²
x ^ 5 = 6 + 60 - 50
x ^ 5 = 16 m
Distance travelled by particle from t = 0 s to t = 3 s,
∆ x = | x³ - x ^ 0 |
∆ x = | 24 - 6 |
∆ x = 18 m
Distance travelled by particle from t = 3 s to t = 5 s,
∆x' = |x5 - x3|
∆x' = |16 - 24|
∆x' = 8 m
Therefore, total distance travelled by particle in 5 s will be -
Distance = ∆ x + ∆ x'
= 18 + 8
= 26 m
Therefore, Total Distance travelled by Particle in 5 s is 26 m.
Thanks ...
Given : the position of a particle moving along x axis varies with time as x= 6 + 12t – 2t² where x is in metres and t is in seconds
To Find : the distance travelled by the particle in first 5 second is
Solution:
x= 6 + 12t – 2t²
v = dx/dt = 12 -4t
=>at t = 3 velocity becomes zero and direction changes
so till 3 sec he moves in one direction and then after that direction gets changed.
t = 0
=> x = 6
t = 3
=> x = 6 + 12(3) - 2(3)²
=> x = 24
Distance covered till 3 sec =24 - 6 = 18 m
t = 5
=> x = 6 + 12(5) - 2(5)²
=> x =16
Distance covered between 3 sec and 5 secs
= | 16 - 24 |
= 8
Hence total distance covered = 18 + 8 = 26 m
distance travelled by the particle in first 5 seconds 26 m
Learn More:
draw the distance time and acceleration time graph corresponding ...
brainly.in/question/17096955
Draw the distance-time and acceleration time graph corresponding ...
brainly.in/question/17102571