A police car is traveling north on a straight road at a constant 20.0 m/sm/s. An SUV traveling north at 30.0 m/sm/s passes the police car. The driver of the SUV suspects he may be exceeding the speed limit, so just as he passes the police car he lets the SUV slow down at a constant 1.90 m/s2m/s2. How much time elapses from when the SUV passes the police car to when the police car passes the SUV

elapsed time is 10.53 sec

Explanation:

Given that;

velocity of the car = 20.0m/s

Initial velocity of the SUV   = 30.0 m/s

SUV slow down at a constant a = 1.90 m/s²

Now, Let Vs  represent the final velocity of the SUV and after time t when the police can cross it and d represent the distance both of the cars covers between the two crossing

so;

d = × t

also d = t - 1/2 at²

so

× t  =   t - 1/2 at²

we substitute

20 × t  =  30 × t - 1/2 × 1.9 × t²

20t  = 30t - 1/2 × 1.9 × t²

1/2 × 1.9 × t² = 30t - 20t

1/2 × 1/9 × t² = 10t

t = 2 × 10 / 1.9               ( as t ≠ 0 )

t = 20 / 1.9

t = 10.53 sec

Therefore, elapsed time is 10.53 sec

insects!

explanation:

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.