A car moving with a speed of 72 km/h is brought to rest in 10 seconds
by applying brakes. Find the magnitude of average retardation due to
braking and distance travelled by car after applying the brakes.
Answers
Answer:
Retardation = -2m/s^2
Distance = 100m
Explanation:
Step 1: convert 72km/hr to meters/second.
(72km/hr)(1000m/1km)(1hr/60min)(1min/60sec)
=20m/s
Step 2: Find your acceleration/Retardation using your kinematics equation
V=Vo+at
V= 0 (zero because it comes to a rest)
Vo = 20m/s
t = 10 seconds
a=-Vo/t
a = -2m/s^2
Step 3: Using Kinematics Equations, find the distance the car will travel with the acceleration found in the given time interval.
Use the final kinematics equation for this
DeltaX = Vot+1/2at^2
DeltaX = (20m/s)(10sec)+1/2(-2m/s^2)(10)^2
DeltaX = (200m) + 1/2(-200m)
DeltaX = 200m-100m
DeltaX = 100m
Answer:
Explanation:
Given : -
- A car is moving with speed 72 km/h on a straight road.
- Suddenly brakes are applied due to which car retards at the rate of 10 m/s?,
To Find : -
- distance travelled by car before coming to rest is
We are given Initial velocity of Car
10 × 2 = 20 m/s
u = 72 km/h
u = 72 × 20/72 m/s
u = 20 m/s
we have
2 a s = v² - u²
substitute all values
2 (-10) ( s ) = (0)² - (20)²
- 20s = - 400
s = 400/ 20
s = 20 m
Hence the answer is 20 m