Question

A speeding motorist traveling 120 km/h passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 10.8

Answer 1

120 km/h = 33.33 m/s   10.8 km/h/s = 3 m/s/s     the motorist will cover a distance of 33.33 m x T(seconds)   the police officer will cover a distance of 1/2 (3) T^2   they will be at the same point when   33.33T = 1.5 T^2   33.33 = 1.5 T   3T = 66.66   T = 22.22 seconds   it will take the officer 22.22 seconds to catch the motorist.     the officer will be moving V=AT = 3m/s x 22.22 seconds = 66.66 m/s   almost 240 km/h (239.976 km/h)

Answer 2

a. It will take 24.6 s for the police officer to reach the speeder

b. The police speed will be 66.42 m / s at this time

Further explanation

Regular straight motion is the motion of objects on a straight track that has a fixed speed

Formula used

$\large{\boxed{\bold{S=v\times\:t}}}$

S = distance = m

v = speed = m / s

t = time = seconds

Straight motion changes regularly are the straight motion of objects that have a fixed acceleration

Formula used

$\large{\boxed{\bold{St=vot+\frac{1}{2}at^2}}}$

V = vo + at

Vt² = vo² + 2as

St = distance on t

vo = initial speed

vt = speed on t

a = acceleration

The full version of the question might be like this:

A speeding motorist traveling 120 km / h passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 10.8 km / h / s (note the mixed units).

(a) How much time will it take for the police officer to reach the speeder, assuming that the speeder maintains a constant speed?

(b) How fast will the police officer be traveling at this time?

A speeding motorist traveling 120 km/h (fixed speed) shows irregular straight motion

Whereas The officer immediately begins pursuit at a constant acceleration of 10.8 km/h/s showing that straight motion changes irregularly

So that the officer can catch A speeding motorist, the distance achieved by both of them is the same and also with the same time

• 1. the distance achieved by the speeding motorist

v = 120 km / h = 33.3 m / s

S1 = v. t

S1 = 33.3t

• 2. the distance reached by the officer

vo = 0

a = 10 km / h / s = 2.7 m / s²

S2 = vot + 1/2 at²

S2 = 0 + 1/2 .2.7t²

S2 = 1.35t²

• a. The distance traveled when the two meets are the same, so the time of both can be determined

S1 = S2

33.3t = 1.35t²

t = 24.6 s

• b. The officer speed

Vt = vo + at

Vt = 0 + 2.7.24.6

Vt = 66.42 m / s

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Keywords: the officer, a speeding motorist

, fixed acceleration, fixed speed