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# Exercise physics need help?

A car start from a standstill with uniform acceleration of a (m/s^2). After a time ΔT = 7.8 s a bullet is shot that can be supposed to be in a costant speed of v = 97.0 m/s, that is the minimum speed needed to hit the car. Find acceleration a. Solution: 6.22.

Thanks

### 2 Answers

- NCSLv 77 months agoFavorite Answer
Since the car is accelerating uniformly, eventually it will reach 97.0 m/s and keep getting faster. At that point it will no longer be reachable by the bullet.

Let's say the bullet is in the air for time "t." If the bullet is just able to reach the car, impact occurs when they have travelled the same distance:

½a(t + 7.8s)² = 97.0m/s * t

I've already claimed that the car must be going 97.0 m/s:

v = 97.0 m/s = a*(t + 7.8s)

so

a = 97.0m/s / (t + 7.8s)

substitute for a in the first equation:

½ * (97.0m/s / (t + 7.8s)) * (t + 7.8s)² = 97.0m/s * t

½ * 97.0m/s * (t + 7.8s) = 97.0m/s * t

½ * (t + 7.8s) = t

t + 7.8s = 2t

7.8 s = t

and now you can solve for the acceleration:

a = 97.0m/s / (7.8 + 7.8)m/s

a = 6.22 m/s

Hope this helps!

- Anonymous7 months ago
So what is your question jackass?

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You're welcome. Thanks for awarding!