Diagram? There is no diagram given.
a.


Cannot solve b without the diagram.
Answer:
cos(θ) = (√33)/7
Step-by-step explanation:
The relevant relation is ...
cos(θ) = √(1 -sin²(θ))
cos(θ) = √(1 -(4/7)²) = √(1 -16/49) = √(33/49)
cos(θ) = (√33)/7
Here we know that the initial velocity of the car is given by:

And the final velocity of the car is given by:

It took 3 seconds to attain the final velocity, so we have 
Therefore, the acceleration can be obtained by:


Plugging the values of the initial, final velocity and the time, we get:

So the acceleration of the car is given by:

Now we need to find the direction of the average acceleration of the car:

Here, x and y are the coefficients of the 'x' and 'y' components of the vector:

Therefore, the direction of the average acceleration of the car is
.
Answer:
40%
Step-by-step explanation:
40 is 40% of 100, you don't need a calculator for that, so, it's a 40% increase
hope this helps:)