Question

An unmarked police car traveling a constant 80.0 km/h is passed by a speeder traveling 140 km/h . Precisely 2.50 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.40 m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Answer 1

Answer:18.55 s

Step-by-step explanation:

Given

Police car velocity
`=80 km/h\approx 38.889 m/s`

speeder velocity
`=140 km/h\approx 22.22 m/s`

After 2.5 s Police car steps on accelerator to produce an acceleration of
`2.40 m/s^2`

In 5 sec speeder has traveled an additional distance of

=(38.889-22.22)2.5

=41.67 m

Therefore Police needs to travel a distance of 41.67 + distance traveled by speeder in next t sec

Distance traveled by speeder is s

`s=38.889* t`---1

By police car

`s+41.67=22.22* t+(2.4* t^2)/(2)`---2

Subtract 1 from 2

`41.67=22.22t+1.2t^2-38.889t`

`1.2t^2-16.67-41.67=0`

t=16.05 s

Thus a total of 2.5+16.054=18.55 s