# Motion with constant velocity – problems and solutions

Solved Problems in Linear Motion – Constant velocity

1. A car travels at a constant 10 m/s. Determine distance after 10 seconds and 60 seconds.

Solution

Constant speed 10 meters/second means car travels 10 meters every 1 second.

After 2 seconds, the car travels 20 meters,

After 5 seconds, the car travels 50 meters,

__After 10 seconds, the car travels 100 meters__,

__After 60 seconds, the car travels 600 meters__.

2. A car travels along a straight road at constant 72 km/h. Determine the car’s distance after 2 minutes and 5 minutes.

Solution

72 km/h = (72)(1000 meters) / 3600 seconds = 72,000 / 3600 seconds = 20 meters/second.

The constant speed at 20 meters/second means car travels 20 meters every 1 second.

__After 120 seconds or 2 minutes, car travels 20 meters x 120 = 2400 meters__,

__After 300 seconds or 5 minutes, car travels 20 meters x 300 = 6000 meters__.

3. A body travels along a straight road for 100 meters in 50 seconds. Determine the speed of the body.

Solution

__100 meters / 50 seconds = 10 meters / 5 seconds = 2 meters/second__.

4. Determine speed according to the diagram below….

Solution

Speed = Distance / time elapsed

__Speed = 2 meters / 1 second = 4 meters / 2 seconds = 6 meters / 3 seconds = 8 meters / 4 seconds = 2 meters/second__.

5. Cars A and B approach each other on parallel tracks. When the distance between the two cars is 100 meters, car A moves at a constant speed of 10 m/s, car B moves at a constant speed of 40 m/s. Determine (a) the distance of car A before passing car B (b) time interval before car B passing car A.

Solution

Car A moving with a constant speed at 10 meters/second, means car A moves as far as 10 meters every 1 second. After 2 seconds, A car move as far as 20 meters.

Car B moves with a constant speed at 40 meters/second, means car B moves as far as 40 meters every 1 second. After 2 seconds, car B moves as far as 80 meters.

20 meters + 80 meters = 100 meters.

(a) The distance of car A before passing car B is 20 meters. The distance of car B before passing car A is 80 meters.

(b) Time interval of car B before passing car A is 2 seconds. Time interval of car A before passing car B is 2 seconds

5. If the speedometer of a car shows 108 km/h, determine the distance traveled by car in one minute.

Solution :

The speedometer is an instrument to measure speed. The speed of a car is 108 km/hour.

108 km / h = (108) (1000 meters) / 3600 seconds = 30 meters/second.

1 minute = 60 seconds

The speed of the car 30 meters/second means the car moves as far as 30 meters in 1 second.

After 1 second, the car moves as far as 1 x 30 meters = 30 meters.

After 2 seconds, the car moves as far as 2 x 30 meters = 60 meters.

After 60 seconds, the car moves as far as 60 x 30 meters = 1800 meters.

6. Tom throws a ball straight to Andrew. Tom and Andrew are separated as far as 10.08 meters. The ball is thrown horizontally and moves at 20 m/s (ignore gravity). Andrew hits the ball 4.00 x 10^{-3} seconds after the ball was thrown. If the hitter moves at a constant speed of 5.00 m/s, the ball is hit by the hitter after the hitter moves as far as…

__Known :__

The distance between Tom and Andrew = 10.08 meters

Ball’s speed (v) = 20 m/s

The time interval (t) = 4 x 10^{-3} seconds = 0.004 seconds

Hitter’s speed (v) = 5 m / s

__Wanted:__ The ball is hit by hitter after the ball moves as far as…

__Solution :__

Ball’s distance :

s_{1} = v t = (20) (0.004) = 0.08 meters

Hitter’s distance :

s_{2} = v t = 5 t

Ball’s distance + hitter’s distance = distance between Tom and Andrew.

0.08 + 5 t = 10.08

5 t = 10.08 – 0.08

5 t = 10

t = 10/5

t = 2 seconds

Hitter’s distance :

s_{2} = v t = 5 t = (5) (2) = 10 meters

7. A hunter with his car is chasing a deer. The car moves at 72 km/h and the deer run at speeds of 64.8 km/h. When the distance between the car and the deer is 2012 meters, the hunter fired his shotgun. Bullets out of the gun at 200 m/s. Determine the time interval of the deer getting shot.

A. 0.5 s

B. 1 s

C. 1.25 s

D. 1.5 s

__Known :__

Speed of car (v_{b}) = 72 km/h = (72)(1000 m) / 3600 s = 20 m/s

Speed of deer (v_{r}) = 64.8 km/h = (64.8)(1000 m) / 3600 s = 64800 m / 3600 s = 18 m/s

When the bullet is fired, the distance between the car and the deer (s) = 202 meters

Speed of fire (v_{p}) = 20 m/s + 200 m/s = 220 m/s

Weapons held by hunters who are in a car that moves with a speed of 20 m/s so that the speed of car is also added to the speed of the bullet.

__Wanted:__ Determine the time interval of the deer getting shot

__Solution :__

Think of cars and deer moving at a constant velocity.

Equation : v = s / t or s = v t

*v = speed, s = distance, t = time interval*

Distance = 202 + X_{r} = 202 + v_{r} t = 202 + 18 t

Distance = Y_{p} = v_{p} t = 220 t

Distance traveled by deer = distance traveled by bullet

202 + 18 t = 220 t

202 = 220 t – 18 t

202 = 202 t

t = 202/202

t = 1 second

The correct answer is B.

### Ebook PDF uniform linear motion problems and solutions

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