Question

PLS HELP I NEED A COMPLETE ANSWER AND PROCEDURE FOR THIS, IT'S TIMED

A motorcyclist heading east through a small town accelerates at a constant 4.0 m/s² after he leaves the city limits. At time t=0 he is 5.0 m east of the city-limits signpost while he moves east at 15 m/s. Where is he when his speed is 25 m/s?

Answer 1

Answer:

The motorcyclist is 55 miles east of the small town.

Explanation:

Motion With Constant Acceleration

It's a type of motion where the velocity of an object changes uniformly in time.

The equation that rules the change of velocities is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Solving [1] for t and substituting into [2] we get the following equation:

$V_f^2=V_o^2+2aX$

The motorcyclist has an acceleration of a=4\ m/s^2 and an initial distance of 5 m where he travels at vo=15 m/s.  It's required to calculate the distance when the speed is vf=25 m/s.

Solving the last equation for X:

$\displaystyle X=\frac{V_f^2-V_o^2}{2a}$

Substituting:

$\displaystyle X=\frac{25^2-15^2}{2*4}$

$\displaystyle X=\frac{625-225}{8}$

X = 50 m

Adding the initial distance:

The motorcyclist is 55 miles east of the small town.