the displacement of a particle moving in a straight line is described by the relation s= 6+12t-2t^2,here s is in meter & t is in second. The distance covered by the particle in first 5sec is ????
Answers
Answer:
The displacement of a particle moving in a straight line is described by the relation, s=6+12t−2t².Here,s is in meters and t in seconds. What is the distance covered by a particle for the first 5s?
Originally Answered: The displacement of a particle moving in a straight line is described by the relation, s=6+12t-2t^2 . Here, ‘s’ is in meters and ‘t’ in seconds. What is the distance covered by a particle for the first 5 ‘s’’?
Explanation:
First, let us find whether the particle reverses its direction of motion during the specified time.
S=6+12t−2t²
v=ds/dt =12–4t
Equating the velocity v with zero (12–4t=0) yields t=3s. This means particle travels for 3 seconds in forward direction and for the next 2 second in backward direction. In order to find the distance in 5 second, we have to find the magnitude of displacement in first 3 second and then 3 second to 5 second and then add the two.
Displacement in first 3s:
s(t=3s)−s(t=0)=(6+12×3−2×9)−6=18ms (assuming SI Unit)
Displacement during 3 to 5 second:
s(t=5s)−s(t=3s)=(6+12×5–2×25)−(6+12×3−2×9)=−8m
Ignoring the negative sign and adding the two displacements we have the total distance
18+8=26m
Alternatively, we can find initial velocity u by putting t = 0 in equation (1). This comes out to be 12 m/s. Thus u = 12 m/s.
Displacement in first 3 s:
s=ut+1/2 at 2=12×3+1/2 (−4)×9=18m
Displacement in next 2s (notice that particle starts its backward journey with u=0
s=ut+1/2 at 2=0−1/2(−4)×4=−8m