Question

A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m. Find the acceleration, assuming it to be constant as a fraction of the free-fall acceleration. The acceleration due to gravity is 9.81 m/s 2 . Answer in units of g

Answer 1

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2$

Converting to m/s²

$a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

$\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g$

Hence, acceleration of the car is -0.64525g