give me the answer of the question given in the above image
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we should find physical quantites for the given following data
i mean fundamental physical quantity
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Given
x = 3 + 4 t + 3 t², x in meters. t in sec.
The particle is moving in a straight line only.
i) Position is given by x.
x(t= 2sec) = 3 + 4*2+ 3*2² = 23 meters
2) Displacement
Position at t=0, is clearly x(t=0) = 3 meters.
Position at t = 2s, is 23 m.
Displacement = 23 - 3 = 20 m
3) velocity: magnitude and direction..
v = dx/dt = 4 + 6 t
v(t=2 s) = 4 + 6 * 2 = 16 m/sec.
Direction is in the positive x direction.
4) acceleration: magnitude and direction.
a = dv/dt = dx²/dt² = 6 m/sec².
Direction of acceleration is in the positive x direction.
x = 3 + 4 t + 3 t², x in meters. t in sec.
The particle is moving in a straight line only.
i) Position is given by x.
x(t= 2sec) = 3 + 4*2+ 3*2² = 23 meters
2) Displacement
Position at t=0, is clearly x(t=0) = 3 meters.
Position at t = 2s, is 23 m.
Displacement = 23 - 3 = 20 m
3) velocity: magnitude and direction..
v = dx/dt = 4 + 6 t
v(t=2 s) = 4 + 6 * 2 = 16 m/sec.
Direction is in the positive x direction.
4) acceleration: magnitude and direction.
a = dv/dt = dx²/dt² = 6 m/sec².
Direction of acceleration is in the positive x direction.
:-)
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