In traveling a distance of 2.3 km between points A and D, a car is driven at 83 km/h from A to B for t seconds and 41 km/h from C to D also for t seconds. If the brakes are applied for 4.4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.

- time t taken for car to travel is 64.57 s

- distance travelled between A and B is 1.4887 km

Explanation:

Given the data in the question;

=  83 km/h = ( 83×1000 / 60×60) =  23.0555 m/s

= 41 km/h = ( 41×1000 / 60×60) =  11.3888 m/s

now, we calculate the acceleration;

a =  (   -   ) / t

we substitute

a =  ( 23.0555 -  11.3888 ) / 4.4

a = 11.6667 / 4.4

a = 2.6515 m/s²

Now equation for displacement from BC

=  t + 1/2.at²

we substitute

=  23.0555×4.4 + 1/2×a×(4.4)²

we substitute -2.6515m/s² for a

=  23.0555×4.4 + 1/2×(-2.6515)×(4.4)²

= 101.4442 - 25.6665

= 75.7792 m

Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m

so

+   +    =  2300 m

we substitute substitute

+  75.7792 m +    =  2300 m

+   = 2300 - 75.7792

+    = 2224.2208 m

so we substitute 23.0555t for  and 11.3888t for    

23.0555t + 11.3888t  = 2224.2208

34.4443t = 2224.2208

t = 2224.2208 / 34.4443

t = 64.57 s

Therefore, time t taken for car to travel is 64.57 s

Distance Between A to B

 = t ×  

we substitute

 = 64.57 s × 23.0555

 = 1488.69 m

 = 1.4887 km

Therefore, distance travelled between A and B is 1.4887 km

1/0.14nm = 10^10/1.4 a lot