QUESTION: If a car at rest accelerated uniformly to a speed of 169 km/hr in 20 second. Calculate the
distance covered by it?
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Answers
Required Answer :
The distance travelled by the car = 469.4 m
Given :
- Initial velocity of the car = 0 m/s
- Final velocity of the car = 169 km/hr
- Time = 20 seconds
To find :
- Distance travelled by the car
Solution :
Firstly, we will convert the final velocity of the car from km/h into m/s. And for that we will multiply the value by 5/18 to convert it into m/s.
⇒ Final velocity = 169 × 5/18
⇒ Final velocity = 845/18
⇒ Final velocity = 46.94 m/s
Using first equation of motion :-
- v = u + at
where,
- v = Final velocity
- u = Initial velocity
- a = Acceleration
- t = Time
⇒ 46.94 = 0 + (a)(20)
⇒ 46.94 = a × 20
⇒ 46.94/20 = a
⇒ 2.347 = a
Acceleration of the body = 2.347 m/s²
Using second equation of motion :-
- s = ut + ½ at²
Substituting the given values :-
⇒ s = (0)(20) + ½ (2.347)(20)²
⇒ s = 0 + ½ × 2.347 × 20 × 20
⇒ s = 2.347 × 10 × 20
⇒ s = 469.4
Therefore, the distance travelled by the car = 469.4 m
Answer:
using motion formula:
v=u+at, t=20sec=20/3600=1/180 hr
169=0+a.(1/180)
a=169×180
using motion formula
s=ut +(1/2)a.t^2
=0×(1/180) + (1/2)×169×180×(1/180)^2
=169/(2×180)=0.469 km