A traffic police officer observes a fast moving car. Due to over speed officer starts his bike, accelerates uniformly to 90 kph in 8 s, and maintaining a constant velocity of 90 kph, overtakes the car 42 s after the car passed him. If he overtakes the carafter 18 s from the instant he starts, find the distance the officer travelled before overtaking and the speed of car.
Answers
Answer :
Given -
- The bike of police accelerates uniformly
- The speed is 90 kph in first 8 s
- The velocity is constant (90 kph)
- He overtakes the car after 18 seconds of starting his bike
- Time taken from the instant when the police saw the car to the instant when he overtakes the car is 42 seconds [Time for which car travelled before the police officer overtakes]
To Find -
- Distance the officer travelled before overtaking ?
- Speed of the car ?
Solution -
Converting the speed of the bike : 90 * 5/18 = 25 m/s.
Now, Firstly we need to Find Distance travelled by the police officer.
Let Total Distance travelled by him as D.
Now, let D1 to be Distance travelled when the bike starting accelerating uniformly in 8 seconds.
And let D2 be the Distance travelled when the velocity is constant before overtaking the car in 18 - 8 = 10 seconds.
So, Finding D1 :
It accelerates uniformly, so let's find its acceleration.
⇒ v = u + at
⇒ 25 = 0 + a*8
⇒ 25 = 8a
⇒ a = 25/8
⇒ Acceleration = 3.125 m/s²
Now,
➜ s = ut + ½at²
➜ s = 0*8 + ½*3.125*8²
➜ s = 0 + 3.125*8*4
➜ s = 100 m
So, D1 = 100 m
Now, Finding D2 :
During this distance, the bike of the police officer was maintaining a constant velocity.
⇒ Speed = Distance/Time
⇒ 25 = Distance/10
⇒ Distance = 25*10
⇒ Distance = 250 m
So, D2 = 250 m
Hence, the total distance travelled by the bike (police officer) is 100 + 250 = 350 m. [D = D1 + D2]
Now, We know that car travelled the same distance.
So, Speed = Distance/Time
⇒ Speed = 350/42 = 8.33 m/s
Hence, Speed of the car is 8.33 m/s.
Answer :
Given that bike was accelerating uniformly to 90 kph in 8 s. Then it maintains a constant velocity of 90 kph. Then he overtakes the car in 42 s after car passed him. He overtakes the car 18 s after his start.
We have to find distance travelled and speed of car.
At first it was moving with uniform acceleration :
Initial velocity (u) = 0 m/s
Time (t) = 8 s
Final speed (v) = 90 kph = 90 × (5/18) = 25 m/s
We will use 1st equation of motion :
→ v = u + at
→ 25 = 0 + a × 8
→ 8a = 25
→ a = 25/8
→ a = 3.125 m/s²
Now using 3rd equation of motion :
→ v² - u² = 2as
→ 25² - 0² = 2 × 3.125 × s
→ 625 - 0 = 6.25s
→ s = 625/6.25
→ s1 = 100 m
Now for the second part :
Speed (v) = 25 m/s
Time (t) = (18 - 8) = 10 s
We know that,
→ Distance = Speed × Time
→ Distance = 25 × 10
→ Distance2 = 250 m
Now total distance :
→ Total distance = s1 + s2
→ Total distance = 100 + 250
→ Total distance = 350 m
Total distance travelled = 350 m [Answer] ✓
Now speed of car :
Distance travelled = 350 m
Time = 42 s
→ Speed = Distance/Time
→ Speed = 350/42
→ Speed = 8.33 m/s
Speed of car = 8.33 m/s [Answer] ✓