Question

MCAT Physics Tutorial: Basic Kinematics Review + Example Problems

Answer 1

Note: on the MCAT you will *not* be given a calculator so you must develop good estimation skills. For these problems we will be rounding g to 10 m/s/s.

Answer 2

\({\bf{Basic~Terminology:}}\)
- displacement: the shortest distance between the endpoint and start point of a moving object (vector)
- distance: the total distance traveled over a period of time (scalar)
- vector: quantity w/ direction + magnitude
- scalar: quantity w/ magnitude only

\({\bf{The~Kinematics~Equations:}}\)
no position given:
vf = vi + at

no acceleration given
Δx = (1/2)Δv * t

final velocity not given:
Δx = v_o * t + (1/2)at^2

time not given:
v^2 = v_o^2 + 2aΔx

\({\bf{Example~Problem~1:}}\) a car accelerates from rest (0 m/s) to [a] m/s in [b] seconds, find the acceleration and distance in terms of [a] and [b]

acceleration = (final velocity - initial velocity)/time = ([a] - 0)/[b] = [a]/[b]

distance = (1/2)Δv*t = (1/2)[a]/[b]

Answer 3

\({\bf{Example~Problem~2:}}\)
A car (let's call it car 1) with constant velocity [a] m/s passes a car (let's call it car 2) at rest. As soon as car 1 passes car 2, car 2 accelerates at a rate of [b] m/s/s. Find the position when the two cars pass again.

Car 1: use the basic distance = velocity * t equation to get Δx = [a]*t
Car 2: use the position kinematics equation Δx = v_o * t + (1/2)at^2 = (1/2)[b]t^2

then solve [a]*t = (1/2)[b]t^2 for t, then simply plug back into either of the car equations to get the position (car 1 is easier)

\({\bf{Example~Problem~3:}}\) a person jumps to a maximum height [a] m. Find the initial velocity and the duration of the whole jump.

Work in the vertical direction only. v^2 = v_o^2 + 2aΔy, plug in -10 m/s/s for the acceleration (since the only thing providing acceleration is gravity) and [a] for Δy. Since this is the maximum height, the final velocity will be 0 and you can then solve for v_o.

From there, you can use vf = v_o + at to find t, then be careful to double the final t value for the whole jump.

Answer 4

\({\bf{Example~Problem~4:}}\) an object is thrown from the top of a building with height [a] m and initial velocity [b] m/s. Find the total time the object spends in the air. Also find the maximum height reached by the ball, using the top of the building as the reference point 0m/

Total flight time:

Δx = v_o * t + (1/2)at^2

plug in the height of the building [a] for Δx, the initial velocity [b], the acceleration due to gravity (-10 m/s/s) and then solve for t.

Maximum height:

v^2 = v_o^2 + 2aΔx

again, take a final velocity of 0 for the object reaching its max height, a as the acceleration due to gravity, and solve for Δx. This will give the height reached by the ball at the top of its trajectory.

Answer 5

let's do one more, this one's a little more challenging

\({\bf{Example~Problem~5:}}\) One object is dropped at rest from the top of a cliff. After one second, a second object is thrown with initial velocity [b]. Find the time and height where the two objects pass.

First find the position/velocity of the first object after 1 second.

Δx = v_o * t + (1/2)at^2

= (1/2)(10) = 5m

vf = vi + at

giving us

vf = (10)*1 = 10 m/s/s

Second, set up the position equations for both objects and solve for t. since we have an initial x position that is not 0 for object 1, decompose delta x into xf - xi.

object 1:
Δx = v_o * t + (1/2)at^2
xf - 5 = (10t) + (1/2)(10)t^2

object 2:
Δx = v_o * t + (1/2)at^2
= [b]t + (1/2)(10)t^2

set them equal to each other and solve for t

then take that t value and plug back into the position equation for either ball and solve for the position

Answer 6

Anyway, that's the end of my tutorial, I hope it was a helpful resource. Source material is the 2nd Edition Barron's Prep book for the new MCAT