18) A man is texting on his phone as he drives hiscar at 40 km/h, he s

1.	18) A man is texting on his phone as he drives hiscar at 40 km/h, he suddenly notices a child iscrossing the road so he rams his brake bringing thecar to stop in 2s. If the car's mass is 4000 kg and heweighs 60 kg, what was the retarding force on thecar?
Answer» Given :- A man is texting on his phone as he drives his car at 40 km/h. He suddenly notices a child is crossing the road so he rams his BRAKE bringing the car to STOP in 2s. Mass₁ = 4000 kg Mass₂ = 60 kg To Find :- The retarding force on the car. Solution :- We know that, u = INITIAL VELOCITY t = Time v = Final velocity f = Force m = Mass Given that, Mass₁ = 4000 kg Mass₂ = 60 kg Total mass (m) = 4000 + 60 = 4060 kg Initial velocity (u) = 40 kmph = 11.11 m/s Final velocity (v) = 0 Time (t) = 2 s By the formula, $\underline{\boxed{\sf Acceleration=\dfrac{(v-u)}{t} }}$ Substituting their values, $\sf a= \dfrac{(0-11.11)}{2}$ $\sf a = -5.55 \ m/s^{2}$ THUS, $\underline{\boxed{\sf Force=Mass \times Acceleration}}$ Substituting their values, $\sf F= 4060 (-5.55)$ $\sf F = -22553.3N \ or \ -2.3 \times 10^{4} N$ Therefore, the retarding force on the car is -22553.3N or -2.3 x 10⁴ N

18) A man is texting on his phone as he drives hiscar at 40 km/h, he suddenly notices a child iscrossing the road so he rams his brake bringing thecar to stop in 2s. If the car's mass is 4000 kg and heweighs 60 kg, what was the retarding force on thecar?

Answer»

A man is texting on his phone as he drives his car at 40 km/h.

He suddenly notices a child is crossing the road so he rams his BRAKE bringing the car to STOP in 2s.

Mass₁ = 4000 kg

Mass₂ = 60 kg

The retarding force on the car.

We know that,

Given that,

Mass₁ = 4000 kg

Mass₂ = 60 kg

Total mass (m) = 4000 + 60 = 4060 kg

Initial velocity (u) = 40 kmph = 11.11 m/s

Final velocity (v) = 0

Time (t) = 2 s

By the formula,

$\underline{\boxed{\sf Acceleration=\dfrac{(v-u)}{t} }}$

Substituting their values,

$\sf a= \dfrac{(0-11.11)}{2}$

$\sf a = -5.55 \ m/s^{2}$

THUS,

$\underline{\boxed{\sf Force=Mass \times Acceleration}}$

Substituting their values,

$\sf F= 4060 (-5.55)$

$\sf F = -22553.3N \ or \ -2.3 \times 10^{4} N$

Therefore, the retarding force on the car is -22553.3N or -2.3 x 10⁴ N