Question 4.21 A particle starts from the origin at t = 0 s with a velocity of and moves in the x-y plane with a constant acceleration of .
(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
(b) What is the speed of the particle at the time?
Chapter Motion In A Plane Page 87
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(a) s=ut+1/2at^2
16=ut+1/2(g)t^2
(b)v^2-u^2= 2as
16=ut+1/2(g)t^2
(b)v^2-u^2= 2as
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(A) Let time taken by particle = t
Given,
X = 16 m
Ux = 0 m/s
And accⁿ in x-axis = 8 m/s²
X = Uxt + 1/2axt²
16 = 0 + 1/2× 8 × t²
t = 2sec .
Similarly , for Y axis ,
Vy = 10 m/s
t = 2sec
accⁿ along y-axis = 2 m/s²
Y = Uyt + 1/2ayt²
= 10×2 + 1/2×2 × 4
= 20 + 4 = 24 m
Hence, at that time y-coordinate = 24m
(B) Vx = Ux + axt
Vx = 0 + 8×2 = 16m/s
Again, Vy = Uy +ayt
= 10 + 2×2 = 14 m/s
Hence, velocity of particle = 16i+14j
Speed of particle = √(16²+14²)
= 21.26 m/s
Given,
X = 16 m
Ux = 0 m/s
And accⁿ in x-axis = 8 m/s²
X = Uxt + 1/2axt²
16 = 0 + 1/2× 8 × t²
t = 2sec .
Similarly , for Y axis ,
Vy = 10 m/s
t = 2sec
accⁿ along y-axis = 2 m/s²
Y = Uyt + 1/2ayt²
= 10×2 + 1/2×2 × 4
= 20 + 4 = 24 m
Hence, at that time y-coordinate = 24m
(B) Vx = Ux + axt
Vx = 0 + 8×2 = 16m/s
Again, Vy = Uy +ayt
= 10 + 2×2 = 14 m/s
Hence, velocity of particle = 16i+14j
Speed of particle = √(16²+14²)
= 21.26 m/s
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