91Ó°ÊÓ

• The velocity of the car = 50 km/h.
• Car decelearate at a rate of .400 m/s²for 50 seconds.

Apply the following equation of motion as:

$\mathit{\text{v}}\mathit{=}\mathit{\text{u}}\mathit{+}\mathit{\text{at}}$

Here, v is the final velocity, u is the initial velocity, a is the acceleration, andt is the time.

Convert the initial speed of the car from 50.0 km/h to m/s as:

$50\text{ k³¾/³ó}=50\text{ }\frac{\text{km}}{\text{h}}×\frac{1000\text{ m}}{1\text{ k³¾}}×\frac{1\text{â€É¾}}{3600\text{ s}}=13.89\text{ m/²õ}$

Substitute 13.89 m/s for u, (-0.4) m/s² for a, and 50 s for t in the above expression, and we get,

$\begin{array}{c}v=13.89\text{ m/²õ}−0.4{\text{ m/²õ}}^{\text{2}}×50\text{ s}\\ =13.89\text{ m/²õ}−20\text{ m/²õ}\\ =−6.11\text{ m/²õ}\end{array}$

Hence, the final velocity is -6.11 m/s.

The result shows that the car moves in the opposite direction. It is unreasonable that on the application of brakes, the car will move in the opposite direction.

The unreasonable premise is that the car traveling at a speed of 13.89 m/s can decelerate at a rate of 0.4 m/s² in 50 seconds.