Motion with constant velocity – problems and solutions

Solved Problems in Linear MotionConstant 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.

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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.

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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….

Constant velocity – problems and solutions 1Solution

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

Constant velocity – problems and solutions 2Car 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 :

s1 = v t = (20) (0.004) = 0.08 meters

Hitter’s distance :

s2 = 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 :

s2 = 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 (vb) = 72 km/h = (72)(1000 m) / 3600 s = 20 m/s

Speed of deer (vr) = 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 (vp) = 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 + Xr = 202 + vr t = 202 + 18 t

Distance = Yp = vp 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.

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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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