Unit 1: Kinematics — PhysiQuest

Unit 1: Kinematics Practice Problems

Problem 1: Constant Velocity Motion

A car travels at a constant speed of 25 m/s.
a) Calculate the distance the car travels in 15 seconds.
b) If the car started from rest and accelerated uniformly to reach 25 m/s over 5 seconds, calculate its acceleration.

Problem 2: Free Fall Motion

An object is dropped from a height of 80 meters.
a) Calculate the time it takes for the object to reach the ground, ignoring air resistance.
b) Determine the speed of the object just before it impacts the ground.

Problem 3: Relative Motion

Two trains are moving towards each other on parallel tracks.
Train A is traveling at 20 m/s, and Train B at 15 m/s. If they start 500 meters apart, calculate:
a) The time until the trains pass each other.
b) The position from Train A’s starting point where they meet.

Problem 4: Uniformly Accelerated Motion

A car accelerates from 0 to 30 m/s in 8 seconds.
a) Calculate the acceleration of the car.
b) Determine the distance covered by the car during this time.
c) If the car continued at 30 m/s for another 10 seconds, calculate the total distance traveled.

Problem 5: Horizontal and Vertical Motion Components

A projectile is launched horizontally from the top of a 60-meter cliff with an initial speed of 10 m/s.
a) Calculate the time it takes for the projectile to reach the ground.
b) Determine the horizontal distance covered by the projectile by the time it reaches the ground.
c) What is the magnitude of the projectile’s final velocity just before it hits the ground?

Problem 6: Displacement and Velocity Graphs

An object’s motion is described by the velocity-time graph below:
a) For the first 3 seconds, the object moves with a constant velocity of 5 m/s. Afterward, it decelerates uniformly to a stop in the next 4 seconds. Sketch the velocity-time graph for this motion.
b) Using the graph, calculate the total distance traveled by the object during this 7-second interval.
c) Determine the object's average speed over the 7 seconds.

Problem 7: Motion in Two Dimensions

A stone is thrown at an angle of 30° above the horizontal with an initial velocity of 20 m/s.
a) Calculate the horizontal and vertical components of the initial velocity.
b) Determine the time the stone spends in the air.
c) Calculate the maximum height reached by the stone.
d) Find the horizontal range of the stone.

Problem 8: Acceleration and Changing Velocity

A motorcycle accelerates from rest to 60 m/s in a straight line with a constant acceleration of 5 m/s².
a) Determine the time taken to reach 60 m/s.
b) Calculate the distance covered during this acceleration.
c) If the motorcycle then decelerates uniformly to a stop over the next 8 seconds, calculate the deceleration and the additional distance covered.

Answer Key for Unit 1: Kinematics Practice Problems

Problem 1: Constant Velocity Motion

a) Distance = velocity×time\text{velocity} \times \text{time}velocity×time
=25 m/s×15 s=375 m= 25 \, \text{m/s} \times 15 \, \text{s} = 375 \, \text{m}=25m/s×15s=375m
b) Acceleration = ΔvΔt=25 m/s−05 s=5 m/s2\frac{\Delta v}{\Delta t} = \frac{25 \, \text{m/s} - 0}{5 \, \text{s}} = 5 \, \text{m/s}^2ΔtΔv=5s25m/s−0=5m/s2

Problem 2: Free Fall Motion

a) Time to reach the ground using y=12gt2y = \frac{1}{2} g t^2y=21gt2:
80=12(9.8)t2⇒t=80×29.8≈4.04 s80 = \frac{1}{2} (9.8) t^2 \Rightarrow t = \sqrt{\frac{80 \times 2}{9.8}} \approx 4.04 \, \text{s}80=21(9.8)t2⇒t=9.880×2≈4.04s
b) Final speed just before impact using v=gtv = gtv=gt:
v=9.8×4.04≈39.6 m/sv = 9.8 \times 4.04 \approx 39.6 \, \text{m/s}v=9.8×4.04≈39.6m/s

Problem 3: Relative Motion

a) Time until trains meet using relative velocity:
Relative velocity = 20+15=35 m/s20 + 15 = 35 \, \text{m/s}20+15=35m/s
Time = 50035≈14.29 s\frac{500}{35} \approx 14.29 \, \text{s}35500≈14.29s
b) Position from Train A's starting point:
Distance covered by Train A = 20×14.29≈285.8 m20 \times 14.29 \approx 285.8 \, \text{m}20×14.29≈285.8m

Problem 4: Uniformly Accelerated Motion

a) Acceleration using a=ΔvΔta = \frac{\Delta v}{\Delta t}a=ΔtΔv:
a=30 m/s8 s=3.75 m/s2a = \frac{30 \, \text{m/s}}{8 \, \text{s}} = 3.75 \, \text{m/s}^2a=8s30m/s=3.75m/s2
b) Distance covered using d=12at2d = \frac{1}{2} a t^2d=21at2:
d=12(3.75)(82)=120 md = \frac{1}{2} (3.75)(8^2) = 120 \, \text{m}d=21(3.75)(82)=120m
c) Total distance with constant velocity after acceleration:
Distance during acceleration = 120 m
Distance at constant speed = 30×10=300 m30 \times 10 = 300 \, \text{m}30×10=300m
Total distance = 120+300=420 m120 + 300 = 420 \, \text{m}120+300=420m

Problem 5: Horizontal and Vertical Motion Components

a) Time to reach ground using y=12gt2y = \frac{1}{2} g t^2y=21gt2:
60=12(9.8)t2⇒t≈3.5 s60 = \frac{1}{2} (9.8) t^2 \Rightarrow t \approx 3.5 \, \text{s}60=21(9.8)t2⇒t≈3.5s
b) Horizontal distance using d=vxtd = v_x td=vxt:
d=10×3.5=35 md = 10 \times 3.5 = 35 \, \text{m}d=10×3.5=35m
c) Final velocity (magnitude) combining horizontal and vertical components:
Vertical velocity at impact = vy=gt=9.8×3.5≈34.3 m/sv_y = gt = 9.8 \times 3.5 \approx 34.3 \, \text{m/s}vy=gt=9.8×3.5≈34.3m/s
Magnitude of final velocity = (10)2+(34.3)2≈35.8 m/s\sqrt{(10)^2 + (34.3)^2} \approx 35.8 \, \text{m/s}(10)2+(34.3)2≈35.8m/s

Problem 6: Displacement and Velocity Graphs

a) The graph is a horizontal line at 5 m/s for the first 3 seconds, then slopes downward to 0 over the next 4 seconds.
b) Total distance traveled:
d=v⋅td = v \cdot td=v⋅t for the first 3 seconds =5×3=15 m= 5 \times 3 = 15 \, \text{m}=5×3=15m
For the deceleration phase, d=12(5)×4=10 md = \frac{1}{2} (5) \times 4 = 10 \, \text{m}d=21(5)×4=10m
Total distance = 15+10=25 m15 + 10 = 25 \, \text{m}15+10=25m
c) Average speed = total distancetotal time=257≈3.57 m/s\frac{\text{total distance}}{\text{total time}} = \frac{25}{7} \approx 3.57 \, \text{m/s}total timetotal distance=725≈3.57m/s

Problem 7: Motion in Two Dimensions

a) Horizontal and vertical components of initial velocity:
vx=20cos⁡(30∘)≈17.32 m/sv_x = 20 \cos(30^\circ) \approx 17.32 \, \text{m/s}vx=20cos(30∘)≈17.32m/s
vy=20sin⁡(30∘)=10 m/sv_y = 20 \sin(30^\circ) = 10 \, \text{m/s}vy=20sin(30∘)=10m/s
b) Time of flight using t=2vyg=2×109.8≈2.04 st = \frac{2v_y}{g} = \frac{2 \times 10}{9.8} \approx 2.04 \, \text{s}t=g2vy=9.82×10≈2.04s
c) Maximum height using h=vy22g=1022×9.8≈5.1 mh = \frac{v_y^2}{2g} = \frac{10^2}{2 \times 9.8} \approx 5.1 \, \text{m}h=2gvy2=2×9.8102≈5.1m
d) Horizontal range using R=vx⋅tR = v_x \cdot tR=vx⋅t:
R=17.32×2.04≈35.34 mR = 17.32 \times 2.04 \approx 35.34 \, \text{m}R=17.32×2.04≈35.34m

Problem 8: Acceleration and Changing Velocity

a) Time to reach 60 m/s:
t=va=605=12 st = \frac{v}{a} = \frac{60}{5} = 12 \, \text{s}t=av=560=12s
b) Distance covered using d=12at2d = \frac{1}{2} a t^2d=21at2:
d=12(5)(122)=360 md = \frac{1}{2} (5)(12^2) = 360 \, \text{m}d=21(5)(122)=360m
c) Deceleration over the next 8 seconds:
a=Δvt=608=7.5 m/s2a = \frac{\Delta v}{t} = \frac{60}{8} = 7.5 \, \text{m/s}^2a=tΔv=860=7.5m/s2
Additional distance = 12×60×8=240 m\frac{1}{2} \times 60 \times 8 = 240 \, \text{m}21×60×8=240m
Total distance = 360+240=600 m360 + 240 = 600 \, \text{m}360+240=600m