Motorbike running at 25 metre per second is slowed down to 5 metre per second in 2.5 seconds.
Calculate:
1)acceleration
2) distance covered in the time it slows down. ans fast
Answers
Answer:
Explanation:
Given,
Initial velocity, u = 25 m/s
Final velocity, v = 5 m/s
Time = 2.5 seconds
To Find,
a. Acceleration
b. Distance travelled
Formula to be used,
1st equation of motion, i.e, v = u + at
3rd equation of motion, i.e, v² - u² = 2as
Solution,
Putting all the values, we get
v = u + at
⇒ 5 = 25 + a × 2.5
⇒ 5 - 25 = 2.5a
⇒ - 20/0.5 = a
⇒ a = - 8 m/s².
As the sign is negative, so deceleration occurred.
Now, the distance travelled,
v² - u² = 2as
⇒ (5)² - (25)² = 2 × (- 8) × s
⇒ 25 - 625 = - 16s
⇒ - 600 = - 16
⇒ - 600/- 16 = s
⇒ s = 37.5 m
Hence, the distance travelled by bus is 37.5 m.
AnsWer :-
❂We're given that, a motor Motorbike running at 25 metre per second i.e initial velocity (u) is 25 m/s and motor bike slowed down to 5 m/s i.e final velocity (v) is 5 m/s. Time interval (t) is 2.5 s.
❂We need to find the acceleration and distance covered in the time it slows down.
❂By using first kinematical equation of motion we can calculate the acceleration of the motor bike :
↠v = u + at
↠5 = 25 + a(2.5)
↠5 - 25 = 2.5a
↠-20 = 2.5a
↠a = -20 ÷ 2.5
↠a = -8 m/s²
- Hence,the acceleration of the motorbike is -8 m/s².
Now, let's find the distance covered in the time it slows down. By using third kinematical equation of motion.
↠v² - u² = 2as
↠(5)² - (25)² = 2 × (-8) × s
↠25 - 625 = 2 × (-8) × s
↠-600 = -16s
↠s = (-600) ÷ (-16)
↠s = 600 ÷ 16
↠s = 37.5 m
- Hence,the distance covered in the time it slows down is 37.5 m.