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प्रश्न
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
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उत्तर १
0.86 m/s2; 54.46° with the direction of velocity
Speed of the cyclist, v = 27 km/h = 7.5 m/s
Radius of the circular turn, r = 80 m
Centripetal acceleration is given as:
`a_c = v^2/r`
=(7.5)^2/80 = 0.7 `"m/s"^2`
The situation is shown in the given figure:
Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between`a_c` and `a_T`is 90°, the resultant acceleration a is given by:
a = `sqrt(a_c^2 + a_T^2)`
=`sqrt((0.7)^2 + (0.5)^2)`
=`sqrt(0.74) = 0.86` `"m/s"^2`
`tan theta = a_c/a_T`
Where `theta` is the angle of the resultant with the direction of velocity
`tan theta = 0.7/0.5 = 1.4`
`theta = tan^(-1)(1.4)`
`=54.46^@`
उत्तर २
Here v = 27 km/h = `27xx5/18` m/s = 7.5 m/s, r = 80 m
and tangential acceleration `a_t = -0.50 "m/s"^2`
:.Centripetal acceleration `a_c = v^2/r = (7.5)^2/80 ms^2` (radially inwards)
Thus as shown in figure two accelerations are mutually perpendicular directions. If `veca` be the resultant acceleration then
|veca| = sqrt(a_t^2+a_c^2) = sqrt((0.5)^2+(0.7)^2) = 0.86 `ms^(-2)`
and tan beta = `a_c/a_t = 0.7/0.5 = 1.4`
`=> beta = tan^(-1) (1.4) = 54.5^@` fromthe direction of negative of the velocity
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