A) The position at t = 2.0 sec is 43.0 m east
B) The position is 55 m east
Explanation:
A)
In order to solve the problem, we take the east direction as positive direction.
We know that:
- at t = 0, the motorcyclist is at a position of ![x_0 = 5.0 m](https://tex.z-dn.net/?f=x_0%20%3D%205.0%20m)
- at t = 0, the initial velocity of the motorcyclist is
east
- The acceleration of the motorcyclist is constant and it is ![a=4.0 m/s^2](https://tex.z-dn.net/?f=a%3D4.0%20m%2Fs%5E2)
Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression
![x(t)=x_0 + v_0t + \frac{1}{2}at^2](https://tex.z-dn.net/?f=x%28t%29%3Dx_0%20%2B%20v_0t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where t is the time.
Substituting t = 2.0 s, we find the position:
![x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m](https://tex.z-dn.net/?f=x%282.0%29%3D%285.0%29%2B%2815%29%282.0%29%2B%5Cfrac%7B1%7D%7B2%7D%284.0%29%282.0%29%5E2%3D43%20m)
B)
The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:
![v(t)=x'(t)=v_0 + at](https://tex.z-dn.net/?f=v%28t%29%3Dx%27%28t%29%3Dv_0%20%2B%20at)
where:
is the initial velocity
is the acceleration
We want to find the time t at which the velocity is
v = 25 m/s
Solving the equation for t,
![t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv-v_0%7D%7Ba%7D%3D%5Cfrac%7B25-15%7D%7B4%7D%3D2.5%20s)
And therefore, the position at t = 2.5 s is:
![x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m](https://tex.z-dn.net/?f=x%282.5s%29%3D5.0%2B%2815.0%29%282.5%29%2B%5Cfrac%7B1%7D%7B2%7D%284%29%282.5%29%5E2%3D55%20m)
Learn more about accelerated motion:
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