Fill in the blanks:

A component of acceleration parallel to the instantaneous velocity changes the particle's speed.

A component of acceleration perpendicular to the instantaneous velocity changes the particle's direction.

An object has an x-component of velocity vx= -2 m/s and its x-component of acceleration is -5 m/s2. Is the particle's speed (a) increasing or (b) decreasing or (c) neither?

The speed is increasing because the acceleration is in the same direction as the velocity. (They are both in the minus-x direction.)

Problem 41 A police car, accelerating uniformly, passes the 10th street off ramp at a speed of 30.0 m/s. At the 12th street ramp, 0.200 km from 10th, its speed is 50.0 m/s. What is the car's acceleration?.

Choose a coordinate system for solving this problem Choose the direction of the coordinate axes, the position of the origin, and the name of the coordinate(s). Write out your choices below, using complete sentences.

I choose x-axis along the freeway, with the origin at the 10th street off-ramp.

Draw a diagram showing your coordinates. Mark on your diagram all of the relevant points described in the problem statement.

Write down all of the given information using relevant algebraic symbols (e.g. vx) and also write down a symbol for any unknowns you are asked to find.

vx (0)=30.0 m/s, vx,f =50.0, ax=?

Write down any relevant mathematical relations that will help you solve the problem. Do not write down relations that may be correct but are not useful for this problem. If you change your mind, just draw a thin line through the irrelevant information.

vx2- vi2=2a(x-xi)

Solve for the car's acceleration. Comment on your result (both its magnitude and its direction).

(50.0 m/s)2 - (30.0 m/s)2 = 2ax(200 m)

ax = 1.60´ 103/400 m/s2 = 4.00 m/s2

Check the significant figures: Each of the given numbers has 3 significant figures, and the subtraction does not reduce the number, (50.02 = 2.50´ 103, 30.02 = 0.900´ 103 , the difference is 2.50´ 103 - 0.900´ 103 = 1.60´ 103 so the answer should also have three figures: ax = 4.00 m/s2 . The value is < g/2, so seems reasonable. The direction is along the positive x-axis, since the value is positive. This is consistent with the car's increasing speed along the positive x-axis.