A car moving along a straight highway with a speed of 108 km/h, is brought to rest! Within a distance of 100 m. How long does it take for the car to stop
Answers
Answer:
100/15 sec
Explanation:
v=u+at
0=30+at. (108km/h=30m/sec)
at=-30
now
s=ut+1/2at²
100=30t+1/2at²
100=t(30+1/2at)
100=t(30+1/2(-30))
100=t(30-15)=t(15)
t=100/15 sec
Given :
Initial Speed (u) = 108 Km/hr
Distance (s) = 100 m
To Find :
How long does it take for the car to stop
Solution :
Using the Kinematic equation for uniformly accelerated motion, We have
v² = u² + 2as
where v = Final velocity; u = initial velocity; a = acceleration; s = Distance
The initial velocity of the car (u) = 108 Km/hr = 30 m/sec
As the car is coming to rest, therefore, its Final velocity (v) = 0
∴ v² = u² - 2as (as retardation is taking place)
⇒ 0² = (30)² - 2×a×100
⇒ 0 = 900 - 200a
⇒ 200a = 900
∴ a = 4.5 m/sec²
Now to find the time taken (t) by the car to stop, we'll use
∴ v = u + at
⇒ 0 = 30 - 4.5t
⇒ 4.5t = 30
∴ t = 6.67 sec
∴ The time taken by the car to stop is 6.67 seconds.