The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.

5 0

3 years ago

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

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Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

8 0

4 years ago