Question

For the instant represented, car has a speed of 100 km/h, which is increasing at the rate of 8 km/h each second. Simultaneously, car B also has a speed of 100 km/h as it rounds the turn and is slowing down at the rate of 8 km/h each second. Determine the acceleration that car appears to have to an observer in car A.

Answer 1

Answer:

acceleration of car B appears at slowing down by 16 km/h each second

Explanation:

As we know by the concept of relative acceleration

$a_{BA} = a_B - a_A$

now we know that

Car A is moving with speed 100 km/h such that its speed is increasing at rate of 8 km/h each second

so acceleration of car A is given as

$a_A = 8 km/h/s$

now car B is also moving at same speed of 100 km/h but its speed is slowing down at rate of 8 km/h each second

so acceleration of car B is negative as it is decelerating

$a_B = - 8 km/h/s$

So here we have from above formula

$a_{BA} = a_B - a_A = - 8 - 8 = -16 km/h/s$

so acceleration of car B appears at slowing down by 16 km/h each second

Answer 2

The acceleration that car B appears to have to an observer in car A is equal to $-16\;km/h/s$.

Given the following data:

• Speed of car A = 100 km/h.

• Rate of increment A = 8 km/h per second.

• Speed of car B = 100 km/h.

• Rate of increment B = 8 km/h per second.

What is relative acceleration?

Relative acceleration can be defined as the acceleration of an observer B with rest to the rest frame of another observer A.

Mathematically, relative acceleration is given by this formula:

$A_{BA}=A_B-A_A$

Substituting the given parameters into the formula, we have;

$A_{BA}=-8-8\\\\A_{BA}=-16\;km/h/s$

Read more on acceleration here: brainly.com/question/24728358