At time 0.25 s the driver position is at 0.45 m if he drives the vehicle at an initial speed of 1. 8 m/s.
At t = 0.25 s
Considering horizontal motion of diver vehicle :-
Initial speed, u = 1.8 m/s
Acceleration , a = 0 m/s²
Displacement, s = ?
Time, t = 0.25 s
the equation of motion, s= ut + 0.5 at²
Substituting all the above given values
s= ut + 0.5 at²
s = 1.8 x 0.25 + 0.5 x 0 x 0.25²
s = 0.45 m
so, the position of vehicle will be 0.45 m at an initial speed of 1. 8 m/s.
the complete question is:
A diver runs horizontally off the end of a 3.0-m-high diving board with an initial speed of 1.8 m/s. Given that the diver's initial position is xi=0 and yi=3.0m, find her x positions at the time t = 0.25 s.
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