Solved Problems in Linear Motion – Constant acceleration
1. A car accelerates from rest to 20 m/s in 10 seconds. Determine the car’s acceleration!
Zgjidhje
I njohur:
Shpejtësia fillestare (vo) = 0 (rest)
Intervali kohor (t) = 10 sekonda
Shpejtësia përfundimtare (vt) = 20 m/s
Shtepi : Acceleration (a)
zgjidhje:
vt vo + në
20 = 0 + (a)(10)
20 = 10 a
a = 20 / 10
a = 2 m/s2
2. A car is decelerating from 30 m/s to rest in 10 seconds. Determine car’s acceleration.
Zgjidhje
I njohur:
Shpejtësia fillestare (vo) = 30 m/s
Shpejtësia përfundimtare (vt) = 0
Intervali kohor (t) = 10 sekonda
Kërkohet: acceleration (a)
zgjidhje:
vt vo + në
0 = 30 + (a)(10)
– 30 = 10 një
a = – 30 / 10
a = -3 m/s2
The negative sign appears because the final shpejtësi is less than the initial velocity.
3. A car starts and accelerates at a constant 4 m/s2 in 1 second. Determine shpejtësi and distance after 10 seconds.
Zgjidhje
(a) Speed
Acceleration 4 m/s2 means speed increase 4 m/s every 1 second. After 2 seconds, car’s speed is 8 m/s. After 10 seconds, car’s speed is 40 m/s.
(b) Distance
I njohur:
Shpejtësia fillestare (vo) = 0
Shpejtësia përfundimtare (vt) = 40 m/s
Përshpejtimi (a) = 4 m/s2
Kërkohet: Distancë
zgjidhje:
s = vo t + ½ në2 = 0 + ½ (4)(102) = (2)(100) = 200 meters
4. A car travels at a constant 10 m/s, then decelerates at a constant 2 m/s2 until rest. Determine time elapsed and car’s distancë before rest.
I njohur:
Shpejtësia fillestare (vo) = 10 m/s
Acceleration (a) = -2 m/s2 (The negative sign appears because the final velocity is less than the initial velocity)
Shpejtësia përfundimtare (vt) = 0 (rest)
Kërkohet: Time interval and distance
zgjidhje:
(a) Time interval (t)
vt vo + në
0 = 10 + (-2)(t)
0 = 10 – 2 t
10 = 2 t
t = 10 / 2 = 5 sekonda
(b) Distance
vt2 vo2 + 2 boshte
0 = 102 + 2(-2) s
0 = 100 – 4 s
100 = 4 s
s = 100 / 4 = 25 meters
5. A car travels at 40 m/s, decelerates at a constant 4 m/s2 until rest. Determine speed and distance after decelerating in 10 seconds!
Zgjidhje
I njohur:
Shpejtësia fillestare (vo) = 40 m/s
Acceleration (a) = -4 m/s2
Intervali kohor (t) = 10 sekonda
Kërkohet: shpejtësia përfundimtare (vt) and distance (s)
zgjidhje:
(a) Final velocity
vt vo + në = 40 + (-4)(10) = 40 – 40 = 0 m/s
0 m/s means car rest.
(b) Distance
s = vo t + ½ në2 = (40)(10) + ½ (-4)(102) = 400 + (-2)(100) = 400 – 200 = 200 meters
6. Determine distance after 10 seconds!

Zgjidhje
Largësia: s = v t = (10-0)(5-0) = (10)(5) = 50 meters
7. Determine distance after 4 seconds!

Zgjidhje
Distance = square area + triangular area
Distance = (8-0)(8-0) + ½ (16-8)(8-0) = (8)(8) + ½ (8)(8) = 64 + 32 = 96 meters
8. Determine car’s distance after 4 seconds!
Zgjidhje

Distance = triangular area = ½ (4-0)(8-0) = ½ (4)(8) = 16 meters
9. A car moves at 90 km/h past a police car that stops by the side of the road. One minute later, the police car chases at 0.8 m / s2. How far the police car reaches the car?
I njohur:
The speed of car (v) = 90 km/hour = 90,000 meters / 3600 seconds = 25 meters/second
Intervali kohor (t) = 1 minutë = 60 sekonda
Acceleration of police’s car (a) = 0.8 m/s2
Initial velocity of police’s car (vo) = 0 m/s
Kërkohet: Distance traveled by police’s car
zgjidhje:
The car moves at a constant velocity. Distance traveled by the car :
Initial distance :
s = v t = (25)(60) = 1500 meters
Final distance :
s = v t = (25)(t)
Total distance = 1500 + 25 t
Police’s car moving at a constant acceleration. Distance traveled by police’s car :
s = vo t + ½ në2 = (0)(t) + ½ (0.8)(t2) = 0 + 0.4 t2 = 0.4 t2
When the police’s car reaches the the car, distance traveled by police’s car same as distance traveled by the car.
Distance traveled by car = distance traveled by police’s car
1500 + 25 t = 0.4 t2
0.4 t2 – 25 t – 1500 = 0
Use quadratic formula :

Distance traveled by police’s car :
s = 0.4 t2 = (0.4)(1002) = (0.4)(10,000) = 4000 metras= 4 km
10. A makinë moves at a constant 24 m/s brakes so that it has a constant deceleration of 0.952 m/s2. Determine the speed of the car after a distance of 250 meteret.
I njohur:
Shpejtësia fillestare (vo) = 24 m/s
përshpejtim (a) = – 0.952 m/s2 (negative signed because deceleration)
Distancë (d) = 250 metras
Kërkohet: Car’s speed after Metër 250s
zgjidhje:
Known : initial speed (vo), nxitim (a), distancë (d), wanted : final speed (vt) so use the equation of vt2 vo2 + 2 a d
vt = final velocity, vo = initial velocity, një = nxitim, d = distancë
vt2 = (24)2 + (2)(-0.952)(250)
vt2 = 576 - 476
vt2 = 100
vt = √100
vt = 10 m/s
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- Distanca dhe zhvendosja
- Shpejtësia mesatare dhe shpejtësia mesatare
- Shpejtësi konstante
- Përshpejtim konstant
- Lëvizja e rënies së lirë
- Lëvizje poshtë në rënie të lirë
- Lëvizja lart e poshtë në rënie të lirë