Question

Motion at Constant Acceleration: A driver is traveling at 22 m/s (50 mph), notices a stopped car 100 m ahead and then slams on the brakes with a deceleration of 4.0 m/s2. What distance did they travel after they slammed on the brakes and came to a comlplete stop?

Answer 1

Answer:

60.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-22^2}{2\times -4}\\\Rightarrow s=60.5\ m$

The distance the car will travel after slamming the brakes is 60.5 m

The car is 100-60.5 = 39.5 away from the stationary car