Solved MRUA Exercises
MRUA exercises solved.
To review carefully put in parentheses. Make no mistake.
1 .- A body moves from rest with constant acceleration of 8 m / s 2. Calculate: a) the speed is within 5 s, b) the distance traveled from rest, in the first 5 s.
Data:
v i = 0 (m / s)
a = 8 (m / s 2)
v f = v i + at = 0 (m / s) + 8 (m / s 2) x 5 (s) = 40 (m / s)
d = v i t + at 2 / 2 = 0 (m / s) x 5 (s) + 8 (m / s 2) x (5 (s)) 2 / 2 = 100 (m)
2 .- The velocity of a vehicle increases smoothly from 15 km / h to 60 km / h in 20 s. Calculate a) the average speed in km / h in m / s, b)
acceleration c) the distance in meters covered during this time.
Remember to turn in km / ham / s must be divided by 3.6.
Data:
v i = 15 (km / h) = 4.167 (m / s)
v f = 60 (km / h) = 16.67 (m / s)
t = 20 (s)
a = (v f – v i) / t = (16.67 (m / s) – 4.167 (m / s)) / 20 (s) = 0.625 (m / s 2)
d = v i t + at 2 / 2 = 4.167 (m / s) x 20 (s) + 0.625 (m / s 2) x (20 (s)) 2 / 2 = 208.34 (m)
3 .- A vehicle traveling at a speed of 15 m / s increases its speed at 1 m / s every second. A) Calculate the distance covered in 6 s. B) If it slows the 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 / s 2)
a) d = v i t + at 2 / 2 = 15 (m / s) x 6 (s) + 1 (m / s 2) x (6 (s)) 2 / 2 = 108 (m)
b) d = v i t + at 2 / 2 = 15 (m / s) x 6 (s) + 1 (m / s 2) x (-6 (s)) 2 / 2 = 72 (m)
t = (v f – v i) / a = (0 (m / s) – 15 (m / s ))/(- 1 (m / s 2)) = 15 (s)
4 .- A car traveling at a speed of 45 km / h, and applied the brakes after 5 s speed has been reduced to 15 km / h. Calculate a) the acceleration b) distance traveled for five seconds.
Data:
v i = 45 (km / h) = 12.5 (m / s)
v f = 15 (km / h) = 4.167 (m / s)
t = 5 (s)
a = (v f – v i) / t = (4.167 (m / s) – 12.5 (m / s)) / 5 (s) = -1.67 (m / s 2)
d = v i t + at 2 / 2 = 12.5 (m / s) x 5 (s) + (-1.67 (m / s 2)) x (5 (s)) 2 / 2 = 41.625 ( m)
Hernan Verdugo Fabiani Professor of Mathematics and Physics 1
5 .- The speed of trains is reduced uniformly from 12 m / s to 5 m / s. Knowing that during this time through a distance of 100 m, calculate a) the acceleration b) the distance traveled to a stop and then assuming the same acceleration.
Data:
v i = 12 (m / s)
v f = 5 (m / s)
d = 100 (m)
a) a = (v f2 – v i2) / 2d = ((5 (m / s)) 2 – (12 (m / s)) 2 / (2 x 100 (m)) = – 0.595 (m / s 2)
b) d = (v f2 – v i2) / 2a = ((0 (m / s)) 2 – (12 (m / s)) 2 / (2 x (-0.595 (m / s 2))) = 121 (m)
6 .- A body that has a velocity of 10 m / s accelerates at 2 m / s 2. Calculate: a) The increase in speed for 1 min. B) The speed at the end of the first minute. C) The average speed during the first minute. D) The space described in 1 minute.
Data:
v i = 10 (m / s)
a = 2 (m / s 2
a) v f – v i = t = 2 (m / s 2) x 60 (s) = 120 (m / s)
b) v f = v i + at = 10 (m / s) + 2 (m / s 2) x 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 i t + at 2 / 2 = 10 (m / s) x 60 (s) + 2 (m / s 2) x (60 (s)) 2 / 2 = 4,200 (m)
7 .- A body that has a velocity of 8 m / s accelerates uniformly so that its march travels 640 m in 40 s. Calculate: a) The average velocity during 40 s. B) The final speed. C) The increased speed at given time. D) acceleration.
Data:
v i = 8 (m / s)
d = 640 (m)
t = 40 (s)
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 x 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 / s 2)
8 .- A car starts from rest with constant acceleration of 5 m / s 2. Calculate the speed it acquires and the space runs after 4 s.
Data:
v i = 0 (m / s)
a = 5 (m / s 2)
t = 4 (s)
v f = 0 (m / s) + 5 (m / s 2) x 4 (s) = 20 (m / s)
d = v i t + at 2 / 2 = 0 (m / s) x 4 (s) + 5 (m / s 2) x (4 (s)) 2 / 2 = 40 (m)
Hernan Verdugo Fabiani Professor of Mathematics and Physics 2
9 .- A body falling 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 is at 6 s after initiating the movement.
Data:
v i = 0 (m / s)
t 1 = 3 (s)
v f = 27 (m / s)
a = (v f – v i) / t = (27 (m / s) – 0 (m / s)) / 3 (s) = 9 (m / s 2)
t 2 = 6 (s)
v f = v i + at = 0 (m / s) + 9 (m / s 2) x 6 (s) = 54 (m)
d = v i t + at 2 / 2 = 0 (m / s) x 6 (s) + 9 (m / s 2) x (6 (s)) 2 / 2 = 162 (m)
10 .- A body starts from rest with constant acceleration leads tours as 250 m, its velocity is 80 m / s. Calculate the acceleration.
Data:
v i = 0 (m / s)
d = 250 (m)
v f = 80 (m / s)
a = (v f2 – v i2) / 2d = ((80 (m / s)) 2 – (0 (m / s)) 2) / (2 x 250 (m)) = 12.8 (m / s 2)
11 .- The speed with which a projectile out of the canyon is 600 m / s. Knowing that the barrel length is 150 cm, calculate the average acceleration of the projectile so far out of the canyon.
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.
a = (v f2 – v i2) / 2d = ((600 (m / s)) 2 – (0 (m / s)) 2) / (2 x 1.5 (m)) = 120,000 (m / s 2)
We talk about average due to acceleration inside the barrel when the projectile is fired, the force behind it is not constant, so acceleration is not.
12 .- A car increases its speed uniformly from 20 m / s to 60 m / s, while drive 200 m. Calculate the acceleration and the time it takes to go from one to another speed.
Data:
v i = 20 (m / s)
v f = 60 (m / s)
d = 200 (m)
a = (v f2 – v i2) / 2d = ((60 (m / s)) 2 – (20 (m / s)) 2) / (2 x 200 (m)) = 8 (m / s 2)
t = (v f – v i) / a = (60 (m / s) – 20 (m / s)) / 8 (m / s 2) = 5 (s)
Hernan Verdugo Fabiani Professor of Mathematics and physics
3
13 .- A plane travels, before takeoff, a distance of 1,800 m in 12 s with constant acceleration. Calculate: a) acceleration, b) the speed at takeoff, c) the distance traveled during the first and the twelfth second.
Data:
d = 1.800 (m)
t = 12 (s)
Assuming that part of the rest:
d = v i t + at 2 / 2, clearing yields:
a = 2 (d – v i t) / t 2 = 2x (1,800 (m) – 0 (m / s) x 12 (s)) / (12 (s)) 2 = 25 (m / s)
b) v f = v i + at = 0 (m / s) + 25 (m / s 2) x 12 (s) = 300 (m / s)
c) position in one second:
d = v i t + at 2 / 2 = 0 (m / s) x 1 (s) + 25 (m / s 2) x (1 (s)) 2 / 2 = 12.5 (m)
position through twelve second:
d = v i t + at 2 / 2 = 0 (m / s) x 12 (s) + 25 (m / s 2) x (12 (s)) 2 / 2 = 1,800 (m)
distance between the first and the twelfth second:
d = 1.800 (m) – 12.5 (m) = 1.787,5 (m)
14 .- A train has a velocity of 60 km / h stops and in 44 s, it stops. 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)
a = (v f – v i) / t = (0 (m / s) – 16.67 (m / s)) / 44 (s) = -0.379 (m / s 2)
d = v i t + at 2 / 2 = 16.67 (m / s) x 44 (s) + -0.379 (m / s 2) x (44 (s)) 2 / 2 = 366.6 (m)
15 .- A body with a speed of 40 m / s, the decreases uniformly at the rate of 5 m / s 2. Calculate: a) the velocity at 6 s, b) the average speed during the 6 s, c) the distance covered in 6 s.
Data:
v i = 40 (m / s)
a = -5 (m / s 2)
t = 6 (s)
v f = v i+ At = 40 (m / s) + -5 (m / s 2) x 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) x 6 (s) = 150 (m)
Hernan Verdugo Fabiani Professor of Mathematics and Physics 4
16 .- When shooting an arrow from a bow, took an acceleration while walking a distance of 0.61 m. If your speed when fired out was 61 m / s, what was the average acceleration applied to the bow?
Data:
v i = 0 (m / s)
d = 0.61 (m)
v f = 61 (m / s)
a = (v f2 – v i2) / 2d = ((61 (m / s)) 2 – (0 (m / s)) 2) / (2 x 0.61 (m)) = 3,050 (m / s 2)
17 .- A spacecraft moves in free space with a constant acceleration of 9.8 m / s 2. A) If part of the point of rest, how long will it take to acquire a speed of one tenth of the speed of light, b) how far will it travel during this time? (speed of light = 3×10 8 m / s)
Data:
a = 9.8 (m / s 2)
v i = 0 (m / s)
v f = v light / 10 = 3×10 8 (m / s) / 10 = 3×10 7 (m / s)
t = (v f – v i) / a = (3×10 7 (m / s) – 0 (m / s)) / 9.8 (m / s 2) = 3.061.224,49 (s) = 35 days 10 h 20 min 24.49 s
d = v i t + at 2 / 2 = 0 (m / s) x 3.061.224,49 (s) + 9.8 (m / s 2) x (3.061.224,49 (s)) 2 / 2 = 4.59 x10 13 (m)
18 .- A jet lands with a speed of 100 m / s and can accelerate to a maximum rate of – 5 m / s 2 when it will stop. A) From the moment he touches the runway, what is the minimum time required before it stops?, b) the aircraft can land on a runway with a length of 0.8 km?
Data:
v i = 100 (m / s)
a = -5 (m / s 2)
v f = 0 (m / s)
t = (v f – v i) / a = (0 (m / s) – 100 (m / s ))/(- 5 (m / s 2)) = 20 (s)
To see if you can land on a runway of 0.8 (km) = 800 (m) we must calculate how far does the information there and then compared with those 800 (m).
d = v i t + at 2 / 2 = 100 (m / s) x 20 (s) + -5 (m / s 2) x (20 (s)) 2 / 2 = 1,000 (m)
It is noted that slowing the rate of -5 (m / s 2) need 1,000 (m) runway, therefore not enough to land on a runway of 800 (m).
Hernan Verdugo Fabiani Professor of Mathematics and Physics 5