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3 years ago

8

How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h

if the cars' front bumpers are initially 119 m apart?

1 answer:

tresset_1 [31]3 years ago

6 0

Answer:

The time taken is $t = 32.5 \ s$

Explanation:

From the question we are told that

The speed of first car is $v_1 = 66.7 \ km/h = 18.3 \ m/s$

The speed of second car is $v_2 = 52.7 \ km/h = 14.64 \ m/s$

The initial distance of separation is $d = 119 \ m$

The distance covered by first car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 0 m/s

and $d_f$ is the final distance covered which is evaluated as $d_f = v_1 * t$

So

$d_t = 0 \ m/s + (v_1 * t )$

$d_t = 0 \ m/s + (18.3 * t )$

The distance covered by second car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 119 m

and $d_f$ is the final distance covered which is evaluated as $d_f = v_2* t$

$d_t = 119 + 14.64 * t$

Given that the two car are now in the same position we have that

$119 + 14.64 * t = 0 + (18.3 * t )$

$t = 32.5 \ s$

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h = height.

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When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

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