Question

Two cars a and b are travelling towards each other on a single-lane road at 24 m/s and 21 m/s respectively. they notice each other when 180 m apart and apply brakes simultaneously. they just succeed in avoiding collision, both stopping simultaneously at the same position. assuming constant retardation for each car, the distance travelled by car a while slowing down is

Answer 1

Stopping distance =u²/2a

U1=24m/s,u2=21m/s retardation=a1=a2=a
Distance between =180

In first case =24²/2a

=576/2a =288/a

In second

=21²/2a

=441/2a

=220.5/a

Distance when the both stop

288/a +220.5/a-180

$( \frac{288 + 220.5}{a} ) - 180$

Answer 2

Given:

Two cars a and b are traveling towards each other at 24m/s and 21m/s respectively.

To Find:

The distance traveled by a car while slowing down is

Solution:

The initial velocity of the car 1 is, $u_{1}$ = 24m/s

The acceleration of the first car, a = 3m/s²

The initial velocity of the second car, $u_{2}$ = 21m/s

The acceleration of the second car, a = 4m/s²

Then, we know that v² = u² - 2as

We find for the first car,

Final velocity = $v_{1}$ = 0m/s

Substituting the values,

-24² = -2(3)×s

∴ $s_{1}$ = 288m

Now, calculating for second car,

Final velocity = $v_{2}$ = 0m/s

Substituting the values

-21² = -2(4)×s

∴$s_{2}$ = 220.5

The distance when the both the cars stop = 288/a + 220.5/a - 180

= (288 + 220.5/a) - 180.

Therefore, the distance traveled by a car while slowing down is (288 + 220.5/a) - 180.