Question

A car is travelling at 15 ms¹ when the speed limit increases and the car accelerates at a constant rate of 3 ms-²

until reaching a top speed of 30 ms-¹.
a Find the distance covered until reaching top speed.
b Once the car is at top speed, there is a set of traffic lights 600 m away. The car maintains 30 ms until it
starts to decelerate at a constant rate of 5 ms2 to come to rest at the lights. Find the time taken from
reaching top speed until it comes to rest at the traffic lights.

Answer 1

a) The distance covered until reaching top speed is 112.5 meters. b) The time taken from reaching top speed until coming to rest at the traffic lights is 6 seconds.

Step-by-step explanation:

a) Distance covered until reaching top speed:

To find the distance covered until reaching top speed, we can use the equation:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Given:

Initial velocity (u): 15 m/s

Final velocity (v): 30 m/s

Acceleration (a): 3 m/s²

Let's solve for the distance (s):

30^2 = 15^2 + 2(3)s

900 = 225 + 6s

6s = 675

s = 112.5 m

Therefore, the distance covered until reaching top speed is 112.5 meters.

b) Time taken from reaching top speed until coming to rest at the traffic lights:

To find the time taken, we can use the equation:

v = u + at

Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given:

Initial velocity (u): 30 m/s

Acceleration (a): -5 m/s² (negative because it's deceleration)

Distance (s): 600 m

Let's solve for the time (t):

0 = 30 + (-5)t

-30 = -5t

t = 6 seconds

Therefore, the time taken from reaching top speed until coming to rest at the traffic lights is 6 seconds.

Answer 2

Step-by-step explanation:

Given:

V₀ = 15 m/s

a₁ = 3 m/s²

V₁ = 30 m/s

S = 600 m

a₂ = - 5 m/s²

V₂ = 0 m/s

___________

t₃ - ?

1)

Vehicle acceleration time:

t₁ = (V₁ - V₀) / a₁ = (30 - 15) / 3 = 15 / 3 = 5 s

2)

Finding the deceleration time of the car:

S = (V₂² - V₁²) / (2*a₂)

2*a₂*S = (V₂² - V₁²)

a₂ = (V₂² - V₁²) / (2*S) = (0² - 30²) / (2*600) = - 0.75 m/s²

t₂ = (V₂ - V₁) / a₂3 = (0 - 30) / (- 0.75) = 40 s

3)

Total time:

t = t₁ + t₂ = 5 + 40 = 45 s