A car is moving with a velocity of 10m/s can be stopped by the application of a constant force F in a distance of 20m.If the velocity of the car is 30m/s,it can be stopped by this force in?
Answers
Explanation:
Answer:
The car can be stopped by this force in 180 m.
Explanation:
Given that,
velocity of car = 10 m/s
distance s = 20 m
Let the mass of the car be m.
Case (I).
Using equation of motion
v^2 =u^2+2asv
2
=u
2
+2as
0=10^2+2a\times200=10
2
+2a×20
a =-\dfrac{100}{40}\ m/s^2a=−
40
100
m/s
2
Negative sign shows the retardation because the speed is decreasing.
For case (II),
Velocity v = 30 m/s
The force is same.
So, The acceleration will be same.
Using equation of motion again
v^2=u^2+2asv
2
=u
2
+2as
0=30^2+2as0=30
2
+2as
s = \dfrac{-30^2}{2\times\dfrac{-100}{40}}s=
2×
40
−100
−30
2
s = 180\ ms=180 m
Hence, The car can be stopped by this force in 180 m.
Concept:
We need to apply the third kinematic equation in the given question. The third kinematic equation is given as- v² = u² + 2as.
Given:
Velocity = 10m/s
Force = F
Distance = 20m
Find:
We need to determine the distance of the car when it is stopped by Force F having an original velocity of 30m/s.
Solution:
The final velocity, v is zero while the acceleration, a is considered negative.
Therefore the equation, v² = u² + 2as becomes-
0 = u² - 2as
Therefore, s = u²/2a where s is the displacement, u is the initial velocity and a is the acceleration.
Thus, it can be considered that distance is directly proportional to the square of the initial velocity.
s ∝ u²
If a car travelling at 10 m/s can be stopped by applying a constant force F over a distance of 20 m, the stopping distance will triple as the velocity increases.
Therefore, 3² = 9times
Therefore, the new stopping distance will become - 9 × the previous distance
= 9 × 20m
= 180m
Thus, the new stopping distance for a car moving with a velocity of 30m/s having force F becomes 180m.
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