Physics question -A driver of a car travelling at 72km/h, applying brakes and accelerates uniformly in opposite direction at the rate of 5m/s2. How long will the car take to come into rest ? How far will the car covered a distance after applying bra
72km/hr = 20 m/s (72*1000/3600)
To find the time it takes for the car to come to a stop, use the equation: v(final)=v(initial) + at
0=20 - 5t
t = 4 seconds.
To find the distance traveled,
use the equation: x = b + vt + (1/2)at^2, where x is x(final), b is x(initial), v is velocity (initial), t is time (seconds), and a is acceleration.
x=0+20(4)+(1/2)(-5)4^2
x=40 m
To find the time it takes for the car to come to rest, we can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s, since the car comes to rest)
u = initial velocity (72 km/h converted to m/s)
a = acceleration (-5 m/s^2, since acceleration is in the opposite direction)
t = time taken
First, let's convert the initial velocity from km/h to m/s:
Initial velocity (u) = 72 km/h = 72 × 1000/3600 m/s ≈ 20 m/s
Using the equation of motion, we have:
0 = 20 + (-5)t
Rearranging the equation to solve for t:
5t = 20
t = 20/5
t = 4 seconds
Therefore, the car will take 4 seconds to come to rest.
To find the distance covered by the car after applying the brakes, we can use the equation:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity
a = acceleration
t = time taken
Plugging in the values:
s = (20 × 4) + (1/2)(-5)(4)^2
s = 80 - 40
s = 40 meters
Therefore, the car will cover a distance of 40 meters after applying the brakes.