Question

A car stopped at a red light starts moving when the light turns green and accelerates at the rate of 2.7 m/s2 for 15 seconds. The driver continues at this speed for two minutes, until he sees the next traffic signal turning red 191 m away. He reacts after 2 seconds and slams on his brakes.

a. What is the car’s acceleration if he stops just as he reaches the light?
b. What is the total distance covered by the car?

Answer 1

Answer:

1 hour

Explanation:

60 knight of the paper work

Answer 2

Given info : A car stopped at a red light starts moving when the light turns green and accelerates at the rate of 2.7 m/s² for 15 seconds. The driver continues at this speed for 2 min , until he sees the next traffic signal turning red 191 m away. He reacts after 2 sec and slams on his brakes.

To find :

1. what is the car's acceleration if he stops just as he reaches the light ?
2. what is the total distance covered by the car ?

solution : case 1 : A car stopped at a red light starts moving when the light turns green and accelerates at the rate of 2.7 m/s² for 15 sec.

initial velocity of car, u = 0 m/s

acceleration of car, a = 2.7 m/s²

time taken, t = 15 sec

velocity after t sec is given by, v = u + at

⇒ v = 0 + 2.7 × 15 = 40.5 m/s

distance covered during t sec is given by, s = $ut+\frac{1}{2}at^2$

s₁ = 0 × 15 + $\frac{1}{2}$ × 2.7 × 15² = 303.75 m

case 2 : driver continues at 40.5 m/s for two minutes.

so, the distance covered in 2 min , s₂ = vt

= 40.5 m/s × 2 × 60 sec

= 4860 m

case 3 :  he sees the next traffic signal turning red, s₃ =  191 m away. He reacts after 2 seconds and slams on his brakes.

distance covered in 2 sec, s = vt = 40.5 m/s × 2 = 81 m

now, the remaining distance, d = 191 - s = 191 - 81 = 110 m

final velocity of the car, v' = 0

using formula, v² = u² + 2as

here, v = v' = 0 m/s , u  = 40.5 m/s and s = d = 110 m

so, 0² = 40.5² + 2a × 110

⇒ a = - $\frac{40.5^2}{220}$ = -7.455 m/s²

therefore the car's acceleration is -7.455 m/s² if he stops just as he reaches the light.

now total distance = s₁ + s₂ + s₃

= 303.75 m + 4860 m + 191 m

= 5354.75 m

therefore the total distance covered by the car is 5354.75 m