A particle moves such that its displacement varies with time according to the relation S = 12t – 2t2. Find distance travelled in 4 s.
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 ?
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First, let us find whether the particle reverses its direction of motion during the specified time.
S=6+12t−2t2
v=dsdt=12–4t … … … (1)
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=18m (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+12at2=12×3+12(−4)×9=18m
Displacement in next 2s (notice that particle starts its backward journey with u=0 ):
˙·٠•●♥ S=UT+12AT2=0−12(−4)×4=−8M ♥●•٠·˙
Answer:12m
Explanation:As put the value of t in the equation.
12×4-2×(4)^2
48-32
16