First we will calculate velocity vector.
To calculate the velocity vector we have to differentiate the Position vector x(t) and y(t), the process of calculation of velocity vector is shown below:
Therefore velocity vector v(t)= d(t5)/dt, d(t3)/dt
=> v(t) = d(t5)/dt, d(t3)/dt = (5t4), (3t2),
Therefore the velocity vector is equals to v(t) = (5t4),(3t2),
Now we will calculate the acceleration vector.
To calculate the acceleration vector we have to differentiate the velocity vector, therefore we get-
a(t) = d(5t4)/dt, d(3t2)/dt = 20t3, 6t
Therefore the acceleration vector is equals to a (t) = 20t3, 6t.
Here acceleration a = t2 + 1,
We know that a = dv / dt,
Therefore dv = a . dt,
Or v = ∫23 a . dt,
So we get-
v = ∫23 (t2 + 1). dt,
v = 2t |23,
v = 2  - 2 ,
v = 2 m / s.
Therefore velocity of the body is 2 m / s.
Here 2 ≤ t ≤ 4.
Now dividing [2, 4] into subintervals of length Δ t and suppose t k be the time in k th subinterval. The amount of consumption of potato during this interval is given by-
R (t k) Δ t million bushels.
The amount consumed for 2 ≤ t ≤ 4 is given by -
∑ R (t k) Δ t million bushels.
Now our Definite Integral for amount of consumption from t = 2 to t = 4 is given by-
∫24 R (t) dt = ∫24 (2.2 + 1.1 t) dt million bushels
After evaluation we get-
7.066 million bushels.
Therefore the consumption is equals to 7.066 million bushels.
Let f (x) be the force required to stretch the spring by 'x' meters from its natural length. Using the Hooke's Law, that is f (x) = k x where ' k ' is constant.
Therefore f (2) = 10 = k. 2,
So k = 5 and f (x) = 5 x for this spring.
Now writing the integral for the work done to apply force 'f' on interval from x = 0 to x = 4.
Now dividing or partitioning the intervals into small subintervals on each of which 'f' is constant so we can apply constant - force formula for work. If ‘xk’ is any Point in k th subinterval, For whole interval 'f' is approximately f (x k) = 5 xk. The amount of work done by force 'f' on interval is 5 x k Δ x, here Δ x denotes the length of the interval. Therefore sum is
∑ f (xk) Δ x = ∑ 5 x k Δ x is used for approximation of the work done by ' f ' from x = 0 to x = 4.
Now integrating we get-
∫04 f (x) dx = ∫04 5 x dx = 5 x2 / 2 | 40 = 40.
Hence amount of work done to stretch the spring is equals to 40.
Acceleration for any length of time 't' adds to the velocity of car we get, v (t) = 5 + ∫ 2.4 u du = 5 + ∫0t 2.4 u . du = 5 + 1.2 t2 mph.
Therefore the distance covered from t = 0 to t = 8 sec is,
∫08| v (t) |dt = ∫08 (5 + 1.2 t2) dt,
=> ∫08 (5 + 1.2 t2) dt = [5 t + 0.4 t3] 08 = 244 mph * seconds = 0.068 mile.
Hence the distance traveled during those 8 seconds is 0.068 mile.
Calculate the net change in velocity, when acceleration is constant velocity change = acceleration * time applied,
Partition time [0, 8] into small subintervals of length Δ t. The acceleration is constant at each sub interval; therefore ‘t k’ is any Point in the k th subinterval, the change in velocity due to acceleration in subinterval is,
a (tk) Δ t mph.
And net change in velocity for 0 ≤ t ≤ 8 is equals to-
∑ a ( tk ) Δ t mph.
Now write the Definite Integral we get
∫ a ( t ) dt,
Net change in velocity = ∫08 2.4 t dt = 1.2 t2 |08 = 76.8 mph.
Hence the velocity of the car when 8 seconds are up is 76.8 mph.