A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?

Distance= 4 miles

Time = 36.3 seconds

Explanation:

80 mph = 178.95 m/s

90 mph = 201.32 m/s

V = u +at

201.32= 0+a(4.5)

201.32/4.5= a

44.738 m/s² = a

Acceleration of the cop car

= 44.738 m/s²

Distance traveled at 4.5seconds

For the cop car

S= ut + ½at²

S= 0(4.5) + ½*44.738*4.5

S= 100.66 meters

Distance traveled at 4.5seconds

For the speeding car

4.5*178.95=805.275

The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal

X/178.95= (704.675+x)/201.32

X-0.89x= 626.37

0.11x= 626.37

X= 5694.3 meters

The time = 5694.3/178.95

Time =31.8 seconds

So the distance they meet

= 5694.3+805.275

= 6499.575 meters

= 4.0 miles

The Time = 4.5+31.8

Time = 36.3 seconds

truck y has a greater velocity than truck x.