So the equation used in this problem is ΔX=V0*T+1/2AT^2 the X is the distance, v0 is initial velocity, T is time, and a is acceleration. So when we plug these values it we get: 108= 0•T+1/2•3•T^2,the 0•t disappears, and the 1/2•3 gets us 1.5, so we have 108=1.5T^2, then we divide 108 by 1.5 which gets us 72=t^2, and we then take the square root and get 8.49=T so the answer is 8.49 seconds.
Answer:
Acceleration
Explanation:
For example, if the acceleration is zero, then the velocity-time graph is a horizontal line (i.e., the slope is zero). If the acceleration is positive, then the line is an upward sloping line (i.e., the slope is positive).
(a) The time it takes for the police officer to catch up to the speeding car is determined as 0.31 s.
(b) The speed of the police officer at the time he catches up to the driver is 136.8 km/h.
<h3>
Time of motion of the police</h3>
The time taken for the police to catch up with the driver is calculated as follows;
v = at
where;
- a is acceleration = 11.8 km/h/s, = 3.278 m/s²
- v is velocity = 135 km/h = 37.5 m/s
t = v/a
t = 37.5/3.278
t = 11.4 seconds
(v1 - v2)t = ¹/₂at² --- (1)
(v1 - v2)t = v1²/2a --- (2)
From (1):
(v1 - 37.5)t = ¹/₂(3.278)t²
(v1 - 37.5)t = 1.639t²
v1 - 37.5 = 1.639t
v1 = 1.639t + 37.5 -----(3)
From (2):
(v1 - 37.5)t = v1²/(2 x 3.278)
(v1 - 37.5)t = 0.153 ----- (4)
solve 3 and 4;
(1.639t + 37.5 - 37.5)t = 0.153
1.639t² = 0.153
t² = 0.0933
t = 0.31 s
<h3>Speed of the police officer</h3>
v1 = 1.639(0.31) + 37.5 = 38 m/s = 136.8 km/h
Learn more about velocity here: brainly.com/question/4931057
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It keeps them back on the seat so they don't fly out the windshield.
