Physics, asked by suni9346, 1 year ago

The following graphs represent velocity – time graph. Identify what kind of motion a particle undergoes in each graph.

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Answered by abhi178
38
(a) here, velocity is directly proportional to time ,
e.g., v = kt
differentiate with respect to t
dv/dt = acceleration = k
Also, v = kt
dx/dt = kt
∫dx = k∫t.dt
x = kt²/2 [ this shows path is parabolic ]
Hence, velocity - time graph shows the motion of object is accelerating. And trajectory of motion must be parabolic.

(b) velocity is Constant with time,
e.g., v = k ,
differentiate with respect to t
dv/dt = acceleration = 0
Again, v = k
dx/dt = k
∫dx = k∫dt
x = kt [ this shows motion of path is st. Line]
Hence, object moves with uniform motion. Also trajectory of motion is straight line .

(c) here shows, particle rest for some time. Let at t = t₀ particle starts to move with constant acceleration a , where trajectory of motion is parabolic .
here, v = k(t - t₀)
dx/dt = k(t - t₀)
x = k(t²/2 - tt₀) [ parabolic path ]
And dv/dt = a = k [ constant acceleration]

(d) velocity - time graph is parabolic .
Hence, v is directly proportional to square of time.
e.g., v = kt²
dx/dt = kt²
dx = kt².dt
x = kt³/3 + Q
Hence, trajectory of motion increases strictly with increases time.
Also, acceleration , a = dv/dt = 2kt [ vary with time ]
acceleration is variable . It increases with time.
Answered by anilsaikia
0

Explanation:

(a) it represents uniform acceleration

(b) it represents uniform velocity

(c) it represents variable velociy

(d) it represents a parabola with increasing slope i.e, it undergoes acceleration

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