Question

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.

Answer 1

- 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;

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

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

now, we calculate the acceleration;

a = (
`U_(BC)` -
`U_(CD)` ) / 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

`S_(BC)` =
`U_(BC)`t + 1/2.at²

we substitute

`S_(BC)` = 23.0555×4.4 + 1/2×a×(4.4)²

we substitute -2.6515m/s² for a

`S_(BC)` = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²

= 101.4442 - 25.6665

`S_(BC)` = 75.7792 m

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

so

`S_(AB)` +
`S_(BC)` +
`S_(CD)` = 2300 m

we substitute substitute

`S_(AB)` + 75.7792 m +
`S_(CD)` = 2300 m

`S_(AB)` +
`S_(CD)` = 2300 - 75.7792

`S_(AB)` +
`S_(CD)` = 2224.2208 m

so we substitute 23.0555t for
`S_(AB)` and 11.3888t for
`S_(CD)`

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

`S_(AB)` = t ×
`U_(AB)`

we substitute

`S_(AB)` = 64.57 s × 23.0555

`S_(AB)` = 1488.69 m

`S_(AB)` = 1.4887 km

Therefore, distance travelled between A and B is 1.4887 km