A particle moves in a straight line with the velocity as shown in the figure. At t = 0, x = - 16 m. The graph cuts time axis at t=24s
A
The maximum value of the position coordinate of the particle is 54 m.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
The maximum value of the position coordinate of the particle is 36 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
The particle is at the position of 36 m at t = 18 s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
The particle is at the position of 36 m at t = 30 s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct options are A The maximum value of the position coordinate of the particle is 54 m. C The particle is at the position of 36 m at t = 18 s D The particle is at the position of 36 m at t = 30 s Maximum value of position coordinate = initial coordinate + area under the graph upto t = 24 s (As upto t = 24 s the displacement of the particle will be positive).
=−16+[(2×10)+(2+62)×(18−10)+1×62×(24−18)] = - 16 + [20 + 32 + 18] = 54 m At time t = 18 s Position = - 16 + Area of graph upto t = 18 s = - 16 + [20 + 32] = 36 m At time t = 30 s Position = - 16 + Area of graph upto t = 30 s =−16+[70−12×6×6]=36m