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 [tex]x_0 = 5.0 m[/tex]
- at t = 0, the initial velocity of the motorcyclist is [tex]v_0 = 15.0 m[/tex] east
- The acceleration of the motorcyclist is constant and it is [tex]a=4.0 m/s^2[/tex]
Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression
[tex]x(t)=x_0 + v_0t + \frac{1}{2}at^2[/tex]
where t is the time.
Substituting t = 2.0 s, we find the position:
[tex]x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m[/tex]
B)
The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:
[tex]v(t)=x'(t)=v_0 + at[/tex]
where:
[tex]v_0=15.0 m/s[/tex] is the initial velocity
[tex]a=4.0 m/s^2[/tex] is the acceleration
We want to find the time t at which the velocity is
v = 25 m/s
Solving the equation for t,
[tex]t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s[/tex]
And therefore, the position at t = 2.5 s is:
[tex]x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m[/tex]
Learn more about accelerated motion:
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