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 :
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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