it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
I think the answer is c) 1
Hope it helps and have a good day
Answers:
a) 
b) 
c) 
Explanation:
<h3>a) Impulse delivered to the ball</h3>
According to the Impulse-Momentum theorem we have the following:
(1)
Where:
is the impulse
is the change in momentum
is the final momentum of the ball with mass 
and final velocity (to the right) 
is the initial momentum of the ball with initial velocity (to the left) 
So:
(2)
(3)
(4)
(5)
<h3>b) Time </h3>
This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 
:
(6)
(7)
Where:
is the acceleration
is the length the ball was compressed
is the time
Finding from (7):
(8)
(9)
(10)
Substituting (10) in (6):
(11)
Finding :
(12)
<h3>c) Force applied to the ball by the bat </h3>
According to Newton's second law of motion, the force 
is proportional to the variation of momentum in time 
:
(13)
(14)
Finally:
From Carnot's theorem, for any engine working between these two temperatures:
efficiency <= (1-tc/th) * 100
Given: tc = 300k (from question assuming it is not 5300 as it seems)
For a, th = 900k, efficiency = (1-300/900) = 70%
For b, th = 500k, efficiency = (1-300/500) = 40%
For c, th = 375k, efficiency = (1-300/375) = 20%
Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid.
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