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F=ma
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according to the v-t graph
V1 = initial velocity = 20cm/s
t1 = initial time = 0
V2 = final velocity = 0cm/s
t2 = final Time = 10
acceleration :- The rate of change in velocity.
(slope of the v-t graph given acceleration )
(slope of the x-t(position-time) graph give velocity)
(area of the a-t graph give distance)
a = ∆v/∆t
a = v2-v1/T2-T1
a = 0-20/10-0
a = -20/10
a = -2cm/s².
F = ma
F = 20g × -2cm/s²
F = -40gcm/s²
F = -40dyne
F = -40 × 10^-5 N
(Here - sign shows foce(i.e friction) is acting opposite of the motion of the ball it also shows force is a vector quantity)
V1 = initial velocity = 20cm/s
t1 = initial time = 0
V2 = final velocity = 0cm/s
t2 = final Time = 10
acceleration :- The rate of change in velocity.
(slope of the v-t graph given acceleration )
(slope of the x-t(position-time) graph give velocity)
(area of the a-t graph give distance)
a = ∆v/∆t
a = v2-v1/T2-T1
a = 0-20/10-0
a = -20/10
a = -2cm/s².
F = ma
F = 20g × -2cm/s²
F = -40gcm/s²
F = -40dyne
F = -40 × 10^-5 N
(Here - sign shows foce(i.e friction) is acting opposite of the motion of the ball it also shows force is a vector quantity)
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