# Motion with constant acceleration – problems and solutions

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!

Solution

Known :

Initial velocity (vo) = 0 (rest)

Time interval (t) = 10 seconds

Final velocity (vt) = 20 m/s

Wanted : Acceleration (a)

Solution :

vt = vo + a t

20 = 0 + (a)(10)

20 = 10 a

a = 20 / 10

a = 2 m/s2

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2. A car is decelerating from 30 m/s to rest in 10 seconds. Determine car’s acceleration.

Solution

Known :

Initial velocity (vo) = 30 m/s

Final velocity (vt) = 0

Time interval (t) = 10 seconds

Wanted : acceleration (a)

Solution :

vt = vo + a t

0 = 30 + (a)(10)

– 30 = 10 a

a = – 30 / 10

a = -3 m/s2

The negative sign appears because the final velocity is less than the initial velocity.

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3. A car starts and accelerates at a constant 4 m/s2 in 1 second. Determine speed and distance after 10 seconds.

Solution

(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

Known :

Initial velocity (vo) = 0

Final velocity (vt) = 40 m/s

Acceleration (a) = 4 m/s2

Wanted : Distance

Solution :

s = vo t + ½ a t2 = 0 + ½ (4)(102) = (2)(100) = 200 meters

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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 distance before rest.

Known :

Initial velocity (vo) = 10 m/s

Acceleration (a) = -2 m/s2 (The negative sign appears because the final velocity is less than the initial velocity)

Final velocity (vt) = 0 (rest)

Wanted : Time interval and distance

Solution :

(a) Time interval (t)

vt = vo + a t

0 = 10 + (-2)(t)

0 = 10 – 2 t

10 = 2 t

t = 10 / 2 = 5 seconds

(b) Distance

vt2 = vo2 + 2 a s

0 = 102 + 2(-2) s

0 = 100 – 4 s

100 = 4 s

s = 100 / 4 = 25 meters

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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!

Solution

Known :

Initial velocity (vo) = 40 m/s

Acceleration (a) = -4 m/s2

Time interval (t) = 10 seconds

Wanted : final velocity (vt) and distance (s)

Solution :

(a) Final velocity

vt = vo + a t = 40 + (-4)(10) = 40 – 40 = 0 m/s

0 m/s means car rest.

(b) Distance

s = vo t + ½ a t2 = (40)(10) + ½ (-4)(102) = 400 + (-2)(100) = 400 – 200 = 200 meters

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6. Determine distance after 10 seconds!

Solution

Distance : s = v t = (10-0)(5-0) = (10)(5) = 50 meters

7. Determine distance after 4 seconds!

Solution

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!

Solution

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?

Known :

The speed of car (v) = 90 km/hour = 90,000 meters / 3600 seconds = 25 meters/second

Time interval (t) = 1 minute = 60 seconds

Acceleration of police’s car (a) = 0.8 m/s2

Initial velocity of police’s car (vo) = 0 m/s

Wanted : Distance traveled by police’s car

Solution :

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 a t2 = (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 meters = 4 km

10. A car 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 meters.

Known :

Initial velocity (vo) = 24 m/s

Acceleration (a) = – 0.952 m/s2 (negative signed because deceleration)

Distance (d) = 250 meters

Wanted : Car’s speed after 250 meters

Solution :

Known : initial speed (vo), acceleration (a), distance (d), wanted : final speed (vt) so use the equation of vt2 = vo2 + 2 a d

vt = final velocity, vo = initial velocity, a = acceleration, d = distance

vt2 = (24)2 + (2)(-0.952)(250)

vt2 = 576 – 476

vt2 = 100

vt = √100

vt = 10 m/s

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1. Distance and displacement
2. Average speed and average velocity
3. Constant velocity
4. Constant acceleration
5. Free fall motion
6. Down motion in free fall
7. Up and down motion in free fall

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