consider the motion in x-direction
= initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m
= acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation
X = t + (0.5) t²
100 = (4.60)
= 21.7 m/s
consider the motion along y-direction
= initial velocity in y-direction = ?
Y = vertical displacement = 0 m
= acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation
Y = t + (0.5) t²
0 = (4.60) + (0.5) (- 9.8) (4.60)²
= 22.54 m/s
initial velocity is given as
= sqrt(()² + ()²)
= sqrt((21.7)² + (22.54)²) = 31.3 m/s
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg
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Answer:
The difference between the velocity graph made walking at a steady rate means that its the same value in time, that means there's no slope on the graph, so its acceleration is 0
On the other hand, if the velocity is increasing with time, the slope of the graph becomes positive, which means that the acceleration of the particle is positive.
Formula for time
t=d/s
so…
t= 48m/4m/s
the two ms cancel each other out and ur left with s
t=12s
