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Marina86 [1]
3 years ago
12

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement

is s(0) = 10 m. Find its position after 4 seconds.
Physics
1 answer:
eimsori [14]3 years ago
5 0

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

a(t)=5+4t-2t^2

Also, a=\dfrac{dv}{dt}

v=\int\limits {a.dt}

v=\int\limits {(5+4t-2t^2).dt}

v=5t+2t^2-\dfrac{2}{3}t^3+c

At t = 0, v(0)=3\ m/s. So, c = 3

v=5t+2t^2-\dfrac{2}{3}t^3+3

Also, v=\dfrac{ds}{dt}, s is the position

s=\int\limits {v.dt}

s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}

s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'

At t = 0, s(0)=10\ m. So, c' = 10

s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10

At t = 4 s

s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.

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