Question

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits. At t=0 he is 5.0m east of the city limit signposts, moving east at 15 m/s. (A) find his position and velocity at t=2.0secs. (B)where is he when his velocity is 25m/secs.

Answer 1

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 $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

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