A car is travelling at a speed of 72km/h, the brakes are applied to produce a uniform acceleration of -1 m/s2. Find the distance it travels after applying brakes so that it completely stops?
Answers
Given :
- Initial velocity of Car, u = 72 km/h
- Acceleration produced due to Breaks, a = -1 m/s²
- Since, Brakes are applied therefore, Final velocity of Car, v = 0
To find :
- Distance covered by Car after applying brakes so that Car completely stops, s = ?
Formulae required :
- Third equation of motion
2 a s = v² - u²
[where a is acceleration, s is distance covered, v is final velocity and u is initial velocity of the body]
Solution :
Converting Initial velocity given in km/h unit into m/s unit
→ u = 72 km/h
→ u = 72 × ( 5 / 18 ) m/s
→ u = 20 m/s
Calculating distance covered by Car using third equation of motion
→ 2 a s = v² - u²
→ 2 ( - 1 ) s = ( 0 )² - ( 20 )²
→ - 2 s = - 400
→ 2 s = 400
→ s = 400 / 2
→ s = 200 m
Therefore,
- Distance covered by Car after applying Brakes is 200 metres.
Answer:
✡ Question ✡
✏ A car is travelling at a speed of 72km/h, the brakes are applied to produce a uniform acceleration of -1 m/s2. Find the distance it travels after applying brakes so that it completely stops?
✡ Given ✡
⚫Initial velocity of a car(u)=72km/h
⚫ Acceleration produced due to break (a) = -1 m/s²
⚫ Final velocity of a car (v) = 0
✡ To Find ✡
✏ Distance it travels after applying brakes so that it completely stops(s) = ?
✡ Formula Used ✡
▶ Third equation of motion:-
⭐ 2as = v²-u² ⭐
( where,
⚫ a = acceleration
⚫ s = distance cover
⚫ v = final velocity )
✡ Solution ✡
➡ We have to convert Initial velocity from km/h unit into m/s unit.
=> u = 72km/h
=> u = 72×(5/18) m/s
=> u = 20 m/s
➡ We have to calculate distance cover by the car using the third equation of motion:-
=> 2as = v²-u²
=> 2(-1)s = (0)²-(20)²
=> -2s = -400
=> s = 400/2
=> s = 200m
Hence,
Distance covered by car after applying brake is 200m.
Explanation: