AP Physics 1 Unit 1 Practice Exam Questions and Answers with Rationales Top Rated A+

AP Physics 1 Unit 1 Practice Exam
Questions and Answers with
Rationales Top Rated A+
Question 1

A student walks 30 m30\text{ m} east and then 40 m40\text{ m} north. What is the
magnitude of the student's displacement?

A. 10 m10\text{ m}
B. 50 m50\text{ m}
C. 70 m70\text{ m}
D. 1200 m1200\text{ m}

Answer: B. 50 m50\text{ m}

Rationale:
The two displacement components are perpendicular, so use the Pythagorean
theorem:

Δr=302+402\Delta r=\sqrt{30^2+40^2} Δr=900+1600=50 m.\Delta
r=\sqrt{900+1600}=50\text{ m}.


Question 2

A car travels at a constant velocity of 18 m/s18\text{ m/s} for 12 s12\text{ s}. How
far does it travel?

A. 1.5 m1.5\text{ m}
B. 30 m30\text{ m}
C. 216 m216\text{ m}
D. 324 m324\text{ m}

Answer: C. 216 m216\text{ m}

Rationale:

,Δx=vt\Delta x=vt Δx=(18)(12)=216 m.\Delta x=(18)(12)=216\text{ m}.


Question 3

An object initially moving at 6 m/s6\text{ m/s} accelerates at 2 m/s22\text{ m/s}^2
for 5 s5\text{ s}. What is its final velocity?

A. 8 m/s8\text{ m/s}
B. 10 m/s10\text{ m/s}
C. 16 m/s16\text{ m/s}
D. 20 m/s20\text{ m/s}

Answer: C. 16 m/s16\text{ m/s}

Rationale:

vf=vi+atv_f=v_i+at vf=6+(2)(5)=16 m/s.v_f=6+(2)(5)=16\text{ m/s}.


Question 4

A velocity-time graph is a straight line that decreases from 20 m/s20\text{ m/s} to
5 m/s5\text{ m/s} over 3 s3\text{ s}. What is the object's acceleration?

A. −15 m/s2-15\text{ m/s}^2
B. −5 m/s2-5\text{ m/s}^2
C. 5 m/s25\text{ m/s}^2
D. 15 m/s215\text{ m/s}^2

Answer: B. −5 m/s2-5\text{ m/s}^2

Rationale:

a=vf−viΔta=\frac{v_f-v_i}{\Delta t} a=5−203=−5 m/s2.a=\frac{5-20}{3}=-5\text{
m/s}^2.


Question 5

An object moves according to

,x(t)=2t2+6t−4.x(t)=2t^2+6t-4.

What is its velocity at t=2 st=2\text{ s}?

A. 8 m/s8\text{ m/s}
B. 10 m/s10\text{ m/s}
C. 14 m/s14\text{ m/s}
D. 16 m/s16\text{ m/s}

Answer: C. 14 m/s14\text{ m/s}

Rationale:

v=dxdt=4t+6.v=\frac{dx}{dt}=4t+6.

At t=2t=2:

v=4(2)+6=14 m/s.v=4(2)+6=14\text{ m/s}.


Question 6

For the object in Question 5, what is its acceleration?

A. 2 m/s22\text{ m/s}^2
B. 4 m/s24\text{ m/s}^2
C. 6 m/s26\text{ m/s}^2
D. 8 m/s28\text{ m/s}^2

Answer: B. 4 m/s24\text{ m/s}^2

Rationale:

v=4t+6v=4t+6

so

a=dvdt=4 m/s2.a=\frac{dv}{dt}=4\text{ m/s}^2.


Question 7

, A ball is thrown vertically upward with an initial velocity of 29.4 m/s29.4\text{
m/s}. Ignoring air resistance, what is its velocity after 2.0 s2.0\text{ s}?

A. 9.8 m/s upward9.8\text{ m/s upward}
B. 9.8 m/s downward9.8\text{ m/s downward}
C. 19.6 m/s upward19.6\text{ m/s upward}
D. 29.4 m/s upward29.4\text{ m/s upward}

Answer: A. 9.8 m/s upward9.8\text{ m/s upward}

Rationale:

v=v0−gtv=v_0-gt v=29.4−(9.8)(2)=9.8 m/s.v=29.4-(9.8)(2)=9.8\text{ m/s}.

The positive value means the ball is still moving upward.



Question 8

At what time does the ball in Question 7 reach its maximum height?

A. 1.0 s1.0\text{ s}
B. 2.0 s2.0\text{ s}
C. 3.0 s3.0\text{ s}
D. 4.0 s4.0\text{ s}

Answer: C. 3.0 s3.0\text{ s}

Rationale:
At maximum height, v=0v=0:

0=29.4−9.8t0=29.4-9.8t t=3.0 s.t=3.0\text{ s}.


Question 9

A position-time graph has a horizontal segment between t=4 st=4\text{ s} and
t=7 st=7\text{ s}. What is the object's velocity during this interval?

A. Positive
B. Negative