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Dmitriy789 [7]
3 years ago
8

An object's acceleration is given by a(t)=a(t)=60t m/s260t m/s2 . if it begins at rest, how far has it gone after 10 seconds?

Physics
1 answer:
Alex Ar [27]3 years ago
6 0
The object's acceleration is given by the expression: a(t) = 60t m/s^2. The distance it traveled after 10 seconds after starting from rest can be obtained by integrating the expression, a(t), twice in order to get the expression for the distance as a function of time. 

The first integral of a(t) is the velocity, v(t), of the object which is then equal to: v(t) = 30t^2. The integral of v(t) is then distance, d(t), of the object which is then equal to: d(t) = 10t^3. With the distance equation available, we simply plug in t = 10 seconds into the equation. It is then determined that the object has traveled 10,000 m or 10 km after 10 seconds starting from rest.    
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Answer:

The direction and magnitude of velocity is 38.65° and 12.005 m/s

Explanation:

Given that,

Mass of car = 1500 kg

Speed of car = 25.0 m/s

Mass of van = 2500 kg

Speed of van =20.0 m/s

We need to calculate the velocity

Using conservation of energy

m_{c}u_{i}+m_{v}u_{i}=(m_{c}+m_{v})v_{f}

1500(25i+0j)+2500(0+20j)=4000(v_{f})

v_{f}=\dfrac{1500\times25}{4000}i+\dfrac{1500\times20}{4000}j

v_{f}=9.375 i+7.5 j

The magnitude of velocity

|v_{f}|=\sqrt{(9.375)^2+(7.5)^2}

|v_{f}|=12.005\ m/s

We need to calculate the direction

\tan\theta=\dfrac{coefficient\ of\ j}{coefficient\ of\ i}

\tan\theta=\dfrac{7.5}{9.375}

\theta=\tan^{-1}0.8

\theta=38.65^{\circ}

Hence, The direction and magnitude of velocity is 38.65° and 12.005 m/s.

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In a situation where a car is rolling back down the hill it is facing up, there could be one or more of several situations at hand.

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