answer the correct option of four questions
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46) option 3
s = u t + 1/2 a t^2
slope = ds/dt = v = u + a t
since a = 4 m/s^2 is +ve, v increases with time and so slope increases with time.
47) option 3.
The slope of position - time graph is same as these lines are parallel to each other... so velocity is constant and same for both cars.
48) option 1
acceleration a = dv / dt = dv/ds * ds/dt = v * dv/ds
From the graph, dv/ds is negative and a constant. Let it be k.
so a = - k v.
v is positive and is decreasing as s increases. So the magnitude of the acceleration is decreasing.
Alternately, a = (v^2 - u^2) / 2 s
v < u from the graph. and v is decreasing as s increases.
So numerator is decreasing and denominator is increasing.
so magnitude of a is decreasing.
49) option 1.
slope du/ds is positive near point 1. so acceleration.
du/ds = 0 at point 2. so no acceleration at that point.
du/ds < 0 at point 3. so deceleration.
s = u t + 1/2 a t^2
slope = ds/dt = v = u + a t
since a = 4 m/s^2 is +ve, v increases with time and so slope increases with time.
47) option 3.
The slope of position - time graph is same as these lines are parallel to each other... so velocity is constant and same for both cars.
48) option 1
acceleration a = dv / dt = dv/ds * ds/dt = v * dv/ds
From the graph, dv/ds is negative and a constant. Let it be k.
so a = - k v.
v is positive and is decreasing as s increases. So the magnitude of the acceleration is decreasing.
Alternately, a = (v^2 - u^2) / 2 s
v < u from the graph. and v is decreasing as s increases.
So numerator is decreasing and denominator is increasing.
so magnitude of a is decreasing.
49) option 1.
slope du/ds is positive near point 1. so acceleration.
du/ds = 0 at point 2. so no acceleration at that point.
du/ds < 0 at point 3. so deceleration.
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