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german
3 years ago
15

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca

r starts from rest and accelerates at 3.9m/s/s, how far away do the cars meet up again?
Physics
1 answer:
4vir4ik [10]3 years ago
3 0

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s  m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t +  1/2 a t²

s =  1/2 a t²   ----------- (2)

compare equation (1)  and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

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b  W_{2.5} =  250 \  J

Explanation:

Generally the force constant is mathematically represented as

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substituting values given in the question

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=>   k  =  80 \ N /m

Generally the workdone  in stretching the spring 3.5 m is mathematically represented as

       W_{3.5} =  \frac{1}{2}  *  k  *  (3.5)^2

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=>    W_{3.5} = 490 \  J

Generally the workdone  in compressing the spring 2.5 m is mathematically represented as

        W_{2.5} =  \frac{1}{2}  *  k  *  (2.5)^2

=>      W_{2.5} =  \frac{1}{2}  *  80 *  (2.5)^2

=>       W_{2.5} =  250 \  J

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