A car moving on a straight road at 18 km /hr increase its velocity to 54 km / hr at a uniform rate in 10 s . Find the acceleration and distance moved in that tie .( hint-convert km /hr into m/sec first )
Initial speed u =18 km /hr =18000 m/ 36000 s =5 m / s
final speed v = 544 km/ hr = _______=_______
(i) Acceleration,a=______________=_________
(ii) Displacement,s=________________=_________________=________________
Answers
Answer:
a= 1m/s² s = 100m
Explanation:
initial velocity of the car (u) = 18km/h
= 18×5/18= 5m/sec
final velocity of the car (v) = 54km/h
= 54×5/18= 15m/sec
time.(t) = 10 seconds
(I)
since v= u+at
=> 15= 5+ a×10
=> a = 1 m/sec²...........................(1)
(ii)
again, v²= u²+2as
=> 15²= 5²+2×1×s
=> 2 s = 225-25 = 200
=> s = 100 metre.........,....(2)
Given
Initial speed = 18 km/h
Final speed = 54 km/h
Time taken = 10 s
To find
i) Acceleration of car
ii) Displacement of car
Solution
Here, initial speed (u) = 18 km/h & final speed (v) = 54 km/h
First let's convert it into m/s
⇒ Initial speed (u) = 18 km/h
= (18 × 1000)/(60 × 60)
= 180/36 m/s
= 5 m/s
⇒ Final speed (v) = 54 km/h
= (54 × 1000)/(60 × 60)
= 540/36
= 15 m/s
⇒ Time (t) = 10 s
Now finding acceleration of car :
We know equation of motion :
➝ Acceleration (a) = (v - u)/t
➝ Acceleration (a) = (15 - 5)/10
➝ Acceleration (a) = 10/10
➝ Acceleration (a) = 1 m/s² (Ans. i)
Now finding displacement of car :
We know equation of motion :
⇒ Displacement (s) = ut + ½at²
⇒ Displacement (s) = 5(10) + ½ (1 × 10²)
⇒ Displacement (s) = 50 + ½(1 × 100)
⇒ Displacement (s) = 50 + ½(100)
⇒ Displacement (s) = 50 + 50
⇒ Displacement (s) = 100 m. (Ans. ii)
Therefore,