Answer:
a) 
b) the motorcycle travels 155 m
Explanation:
Let
, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

where:
is the speed of the motorcycle at time 2
is the velocity of the car (constant)
is the velocity of the car and the motorcycle at time 1
d is the distance between the car and the motorcycle at time 1
x is the distance traveled by the car between time 1 and time 2
Solving the system of equations:
![\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dcar%26motorcycle%5C%5Cx%3Dv_0%5CDelta%7Bt%7D%26x%2Bd%3D%28%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%7D%29%20%5CDelta%7Bt%7D%5Cend%7Barray%7D%5Cright%5D)

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

More force is needed for more mass. Therefore, if the mass is greater and the force is not enough then the object will less likely accelerate
Decrease the amount of force applied
The SI unit of measure for work, as well as
for all other kinds of energy, is the "joule".