Uniformly Accelerated Rectilinear Motion Exercises

Posted on Dec 5, 2024 in Physics

Solved Exercises

Exercise 1
A body moves from rest with a constant acceleration of 8 m/s². Calculate:

1. The speed within 5 seconds.
2. The distance traveled from rest in the first 5 seconds.

Data:
v_i = 0 m/s
a = 8 m/s²
t = 5 s
Solution:
v_f = v_i + at = 0 m/s + 8 m/s² × 5 s = 40 m/s
d = v_it + (at²/2) = 0 m/s × 5 s + (8 m/s² × (5 s)²)/2 = 100 m

Exercise 2
The velocity of a vehicle increases smoothly from 15 km/h to 60 km/h in 20 s. Calculate:

1. The average speed in km/h and m/s.
2. The acceleration.
3. The distance in meters covered during this time.

Remember: To convert from km/h to m/s, divide by 3.6.
Data:
v_i = 15 km/h = 4.17 m/s
v_f = 60 km/h = 16.67 m/s
t = 20 s
Solution:
a = (v_f – v_i)/t = (16.67 m/s – 4.17 m/s)/20 s = 0.625 m/s²
d = v_it + (at²/2) = 4.17 m/s × 20 s + (0.625 m/s² × (20 s)²)/2 = 208.34 m

Exercise 3
A vehicle traveling at a speed of 15 m/s increases its speed by 1 m/s every second.

1. Calculate the distance covered in 6 s.
2. If it slows down at a rate of 1 m/s each second, calculate the distance covered in 6 s and the time it takes to stop.

Data:
v_i = 15 m/s
a = 1 m/s²
t = 6 s
Solution:
a) d = v_it + (at²/2) = 15 m/s × 6 s + (1 m/s² × (6 s)²)/2 = 108 m
b) d = v_it + (at²/2) = 15 m/s × 6 s + (-1 m/s² × (6 s)²)/2 = 72 m
t = (v_f – v_i)/a = (0 m/s – 15 m/s)/(-1 m/s²) = 15 s

Exercise 4
A car traveling at a speed of 45 km/h applies the brakes. After 5 s, the speed has been reduced to 15 km/h. Calculate:

1. The acceleration.
2. The distance traveled for five seconds.

Data:
v_i = 45 km/h = 12.5 m/s
v_f = 15 km/h = 4.17 m/s
t = 5 s
Solution:
a = (v_f – v_i)/t = (4.17 m/s – 12.5 m/s)/5 s = -1.67 m/s²
d = v_it + (at²/2) = 12.5 m/s × 5 s + (-1.67 m/s² × (5 s)²)/2 = 41.63 m

Exercise 5
The speed of a train is reduced uniformly from 12 m/s to 5 m/s. Knowing that during this time it travels a distance of 100 m, calculate:

1. The acceleration.
2. The distance traveled to a stop, assuming the same acceleration.

Data:
v_i = 12 m/s
v_f = 5 m/s
d = 100 m
Solution:
a) a = (v_f² – v_i²)/(2d) = ((5 m/s)² – (12 m/s)²)/(2 × 100 m) = -0.595 m/s²
b) d = (v_f² – v_i²)/(2a) = ((0 m/s)² – (12 m/s)²)/(2 × (-0.595 m/s²)) = 121 m

Exercise 6
A body that has a velocity of 10 m/s accelerates at 2 m/s². Calculate:

1. The increase in speed for 1 min.
2. The speed at the end of the first minute.
3. The average speed during the first minute.
4. The distance covered in 1 minute.

Data:
v_i = 10 m/s
a = 2 m/s²
t = 60 s
Solution:
a) v_f – v_i = at = 2 m/s² × 60 s = 120 m/s
b) v_f = v_i + at = 10 m/s + 2 m/s² × 60 s = 130 m/s
c) v = (v_f + v_i)/2 = (130 m/s + 10 m/s)/2 = 70 m/s
d) d = v_it + (at²/2) = 10 m/s × 60 s + (2 m/s² × (60 s)²)/2 = 4200 m

Exercise 7
A body that has a velocity of 8 m/s accelerates uniformly so that in its motion it travels 640 m in 40 s. Calculate:

1. The average velocity during 40 s.
2. The final speed.
3. The increased speed at the given time.
4. The acceleration.

Data:
v_i = 8 m/s
d = 640 m
t = 40 s
Solution:
a) v = d/t = 640 m/40 s = 16 m/s
b) v = (v_f + v_i)/2, then v_f = 2v – v_i = 2 × 16 m/s – 8 m/s = 24 m/s
c) v_f – v_i = 24 m/s – 8 m/s = 16 m/s
d) a = (v_f – v_i)/t = (24 m/s – 8 m/s)/40 s = 0.4 m/s²

Exercise 8
A car starts from rest with a constant acceleration of 5 m/s². Calculate the speed it acquires and the distance it runs after 4 s.
Data:
v_i = 0 m/s
a = 5 m/s²
t = 4 s
Solution:
v_f = 0 m/s + 5 m/s² × 4 s = 20 m/s
d = v_it + (at²/2) = 0 m/s × 4 s + (5 m/s² × (4 s)²)/2 = 40 m

Exercise 9
A body falls down an inclined plane with constant acceleration from rest. Knowing that after 3 s the speed acquired is 27 m/s, calculate the speed and distance traveled at 6 s after initiating the movement.
Data:
v_i = 0 m/s
t₁ = 3 s
v_f1 = 27 m/s
t₂ = 6 s
Solution:
a = (v_f1 – v_i)/t₁ = (27 m/s – 0 m/s)/3 s = 9 m/s²
v_f2 = v_i + at₂ = 0 m/s + 9 m/s² × 6 s = 54 m/s
d = v_it₂ + (at₂²/2) = 0 m/s × 6 s + (9 m/s² × (6 s)²)/2 = 162 m

Exercise 10
A body starts from rest with constant acceleration and travels 250 m. Its final velocity is 80 m/s. Calculate the acceleration.
Data:
v_i = 0 m/s
d = 250 m
v_f = 80 m/s
Solution:
a = (v_f² – v_i²)/(2d) = ((80 m/s)² – (0 m/s)²)/(2 × 250 m) = 12.8 m/s²

Exercise 11
The speed with which a projectile leaves a cannon is 600 m/s. Knowing that the barrel length is 150 cm, calculate the average acceleration of the projectile inside the cannon.
Data:
v_f = 600 m/s
d = 150 cm = 1.5 m
v_i = 0 m/s (The projectile, before being shot, is at rest).
Solution:
a = (v_f² – v_i²)/(2d) = ((600 m/s)² – (0 m/s)²)/(2 × 1.5 m) = 120,000 m/s²
We talk about average acceleration because inside the barrel, when the projectile is fired, the force behind it is not constant, so the acceleration is not constant either.

Exercise 12
A car increases its speed uniformly from 20 m/s to 60 m/s while driving 200 m. Calculate the acceleration and the time it takes to go from one speed to the other.
Data:
v_i = 20 m/s
v_f = 60 m/s
d = 200 m
Solution:
a = (v_f² – v_i²)/(2d) = ((60 m/s)² – (20 m/s)²)/(2 × 200 m) = 8 m/s²
t = (v_f – v_i)/a = (60 m/s – 20 m/s)/8 m/s² = 5 s

Exercise 13
A plane travels, before takeoff, a distance of 1800 m in 12 s with constant acceleration. Calculate:

1. The acceleration.
2. The speed at takeoff.
3. The distance traveled during the first and the twelfth second.

Data:
d = 1800 m
t = 12 s
v_i = 0 m/s (Assuming it starts from rest)
Solution:
a) a = 2(d – v_it)/t² = 2 × (1800 m – 0 m/s × 12 s)/(12 s)² = 25 m/s²
b) v_f = v_i + at = 0 m/s + 25 m/s² × 12 s = 300 m/s
c) Position in one second:
d₁ = v_it + (at²/2) = 0 m/s × 1 s + (25 m/s² × (1 s)²)/2 = 12.5 m
Position in twelve seconds:
d₁₂ = v_it + (at²/2) = 0 m/s × 12 s + (25 m/s² × (12 s)²)/2 = 1800 m
Distance between the first and the twelfth second:
d = 1800 m – 12.5 m = 1787.5 m

Exercise 14
A train traveling at a velocity of 60 km/h stops in 44 s. Calculate the acceleration and the distance it travels until it stops.
Data:
v_i = 60 km/h = 16.67 m/s
t = 44 s
v_f = 0 m/s
Solution:
a = (v_f – v_i)/t = (0 m/s – 16.67 m/s)/44 s = -0.38 m/s²
d = v_it + (at²/2) = 16.67 m/s × 44 s + (-0.38 m/s² × (44 s)²)/2 = 366.7 m

Exercise 15
A body with a speed of 40 m/s decreases its speed uniformly at a rate of 5 m/s². Calculate:

1. The velocity at 6 s.
2. The average speed during the 6 s.
3. The distance covered in 6 s.

Data:
v_i = 40 m/s
a = -5 m/s²
t = 6 s
Solution:
v_f = v_i + at = 40 m/s + (-5 m/s² × 6 s) = 10 m/s
v = (v_i + v_f)/2 = (40 m/s + 10 m/s)/2 = 25 m/s
d = vt = 25 m/s × 6 s = 150 m

Exercise 16
When shooting an arrow from a bow, it accelerates while traveling a distance of 0.61 m. If its speed when leaving the bow was 61 m/s, what was the average acceleration applied to the arrow?
Data:
v_i = 0 m/s
d = 0.61 m
v_f = 61 m/s
Solution:
a = (v_f² – v_i²)/(2d) = ((61 m/s)² – (0 m/s)²)/(2 × 0.61 m) = 3050 m/s²

Exercise 17
A spacecraft moves in free space with a constant acceleration of 9.8 m/s².

1. If it starts from rest, how long will it take to acquire a speed of one-tenth the speed of light?
2. How far will it travel during this time?

(speed of light = 3 × 10⁸ m/s)
Data:
a = 9.8 m/s²
v_i = 0 m/s
v_f = (speed of light)/10 = (3 × 10⁸ m/s)/10 = 3 × 10⁷ m/s
Solution:
t = (v_f – v_i)/a = (3 × 10⁷ m/s – 0 m/s)/9.8 m/s² = 3.06 × 10⁶ s = 35.4 days
d = v_it + (at²/2) = 0 m/s × 3.06 × 10⁶ s + (9.8 m/s² × (3.06 × 10⁶ s)²)/2 = 4.59 × 10¹³ m

Exercise 18
A jet lands with a speed of 100 m/s and can decelerate at a maximum rate of -5 m/s² until it stops.

1. From the moment it touches the runway, what is the minimum time required before it stops?
2. Can the aircraft land on a runway with a length of 0.8 km?

Data:
v_i = 100 m/s
a = -5 m/s²
v_f = 0 m/s
Solution:
t = (v_f – v_i)/a = (0 m/s – 100 m/s)/(-5 m/s²) = 20 s
To see if it can land on a runway of 0.8 km = 800 m, we must calculate how far it travels during deceleration and compare it with 800 m.
d = v_it + (at²/2) = 100 m/s × 20 s + (-5 m/s² × (20 s)²)/2 = 1000 m
The jet needs 1000 m of runway to stop, therefore, a runway of 800 m is not long enough.