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 (v_{o}) = 0 (rest)

Time interval (t) = 10 seconds

Final velocity (v_{t}) = 20 m/s

__Wanted __: Acceleration (a)

__Solution :__

v_{t} = v_{o} + a t

20 = 0 + (a)(10)

20 = 10 a

a = 20 / 10

a = 2 m/s^{2}

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

Solution

__Known :__

Initial velocity (v_{o}) = 30 m/s

Final velocity (v_{t}) = 0

Time interval (t) = 10 seconds

__Wanted :__ acceleration (a)

__Solution :__

v_{t} = v_{o} + a t

0 = 30 + (a)(10)

– 30 = 10 a

a = – 30 / 10

a = -3 m/s^{2}

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

3. A car starts and accelerates at a constant 4 m/s^{2 }in 1 second. Determine speed and distance after 10 seconds.

Solution

(a) Speed

Acceleration 4 m/s^{2} 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 (v_{o}) = 0

Final velocity (v_{t}) = 40 m/s

Acceleration (a) = 4 m/s^{2}

__Wanted :__ Distance

__Solution :__

s = v_{o} t + ½ a t^{2 }= 0 + ½ (4)(10^{2}) = (2)(100) = 200 meters

4. A car travels at a constant 10 m/s, then decelerates at a constant 2 m/s^{2 }until rest. Determine time elapsed and car’s distance before rest.

__Known :__

Initial velocity (v_{o}) = 10 m/s

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

Final velocity (v_{t}) = 0 (rest)

__Wanted :__ Time interval and distance

__Solution :__

(a) Time interval (t)

v_{t} = v_{o} + a t

0 = 10 + (-2)(t)

0 = 10 – 2 t

10 = 2 t

t = 10 / 2 = 5 seconds

(b) Distance

v_{t}^{2} = v_{o}^{2} + 2 a s

0 = 10^{2} + 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/s^{2 }until rest. Determine speed and distance after decelerating in 10 seconds!

Solution

__Known :__

Initial velocity (v_{o}) = 40 m/s

Acceleration (a) = -4 m/s^{2}

Time interval (t) = 10 seconds

__Wanted :__ final velocity (v_{t}) and distance (s)

__Solution :__

(a) Final velocity

v_{t} = v_{o} + a t = 40 + (-4)(10) = 40 – 40 = 0 m/s

0 m/s means car rest.

(b) Distance

s = v_{o} t + ½ a t^{2} = (40)(10) + ½ (-4)(10^{2}) = 400 + (-2)(100) = 400 – 200 = 200 meters

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/s^{2}. 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/s^{2}

Initial velocity of police’s car (v_{o}) = 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 = v_{o} t + ½ a t^{2 }= (0)(t) + ½ (0.8)(t^{2}) = 0 + 0.4 t^{2 }= 0.4 t^{2 }

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 t^{2}

0.4 t^{2 }– 25 t – 1500 = 0

Use quadratic formula :

__Distance traveled by police’s car :__

s = 0.4 t^{2 }= (0.4)(100^{2}) = (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/s^{2}. Determine the speed of the car after a distance of 250 meters.

__Known :__

Initial velocity (v_{o}) = 24 m/s

Acceleration (a) = – 0.952 m/s^{2 }(negative signed because deceleration)

Distance (d) = 250 meters

__Wanted :__ Car’s speed after 250 meters

__Solution :__

Known : initial speed (v_{o}), acceleration (a), distance (d), wanted : final speed (v_{t}) so use the equation of v_{t}^{2} = v_{o}^{2} + 2 a d

v_{t }= final velocity, v_{o} = initial velocity, a = acceleration, d = distance

v_{t}^{2} = (24)^{2} + (2)(-0.952)(250)

v_{t}^{2} = 576 – 476

v_{t}^{2} = 100

v_{t }= √100

v_{t }= 10 m/s

[wpdm_package id=’507′]

[wpdm_package id=’517′]

- Distance and displacement
- Average speed and average velocity
- Constant velocity
- Constant acceleration
- Free fall motion
- Down motion in free fall
- Up and down motion in free fall