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Find the distance & acceleration from time 0 - 2 seconds, 2 - 10 seconds, 10 - 12 seconds, 12 - 16 seconds. Also, tell the type of motion in all the regions.
"
Hello Dania,
For this question, first, we mark the point and their coordinates.
In 0−2 seconds (Region OA)
Acceleration = Slope of OA Displacement = Area under velocity−time(v−t) graph
= x2−x1y2−y1 = Area of triangle AOP
= 2−08−0 scwwcwcece ddceede
= 21×OA×OP
=28 = 21×2×8
= 4 m/s2 = 8 m
In 2−10 seconds (Region AB)
Acceleration = Slope of AB Displacement = Area under velocity−time(v−t) graph
= x2−x1y2−y1 = Area of rectangle ABQP
= 2−08− 8 scwwcwcece ddceede
= Length(AP) × Breadth(AB)
=20 = 8 ×8
= 4 m/s2 = 8 m
Acceleration = Slope of BC Displacement = Area under velocity−time(v−t) graph
=x2−x1y2−y1 =21 × (sum of parallel sides) × Height
= 12−104−8 scwwcwcece ddceede
= 21× (BQ + CR) × QR
=2-4 = 21× (4 + 8) × 2
=−2 m/s2 =12 m
Acceleration = Slope of CD Displacement = Area under velocity−time(v−t) graph
= x2−x1y2−y1 = Area of rectangle CDSR
=16−124−4 =21 × Length(CD) × Breadth(CR)
=40 = 4 × 4
= 0 m/s2 = 16 m
Motion in region OA = Uniformly Accelerated
Motion in region AB = Uniform Motion (No Acceleration)
Motion in region BC = Uniformly Retarded
Motion in region CD = Uniform Motion (No Acceleration)
Total Displacement = Displacement in OA + Displacement in AB
+Displacement in BC +Displacement in CD
= 8 + 64 + 12 + 16
=100
Here we find the displacement not the direction because It's a velocity-time graph.
To find a distance we should have a speed-time graph.
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