Question

. A driver in a car traveling at a speed of 21.8 m/s sees a cat 101 m away on the road

How long will it take for the car to accelerate uniformly to a stop in exactly 99 ml
- A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for
3.2 km in 3.5 min. How fast (in m/s) is the car moving after this time?

Answer 1

1. The time taken for the car to accelerate uniformly to a stop in exactly 99 m is 9.1 s

2. The speed (in m/s) of the car in 3.5 mins is 24.1 m/s

How calculate the time and speed of the car?

1. The time taken for the car to accelerate uniformly to a stop in exactly 99 m can be calculated as follow:

First, we shall calculate the deceleration of the car. This is shown below:

• Initial velocity (u) = 21.8 m/s
• Distance (s) = 99 m
• Final velocity (v) = 0 m/s
• Deceleration (a) = ?

v² = u² + 2as

0² = 21.8² + (2 × a × 99)

0 = 475.24 + 198a

0 - 475.24 = 198a

-475.24 = 198a

Divide both side by 198

a = -475.24 / 198

a = -2.4 m/s²

Finally, we shall obtain the time. Details below:

• Initial velocity (u) = 21.8 m/s
• Final velocity (v) = 0 m/s
• Deceleration (a) = -2.4 m/s²
• Time (t) =?

`t = (v\ -\ u)/(a) \\\\t = (0\ -\ 21.8)/(-2.35) \\\\t = (-21.8)/(-2.4) \\\\t = 9.1\ s`

2. The speed (in m/s) of the car in 3.5 mins can be calculated as follow:

• Initial speed (u) = 6.4 m/s
• Distance (s) = 3.2 km = 3.2 × 1000 = 3200 m
• Time (t) = 3.5 min = 3.5 × 60 = 210 s
• Final speed (v) = ?

`s = ((u\ +\ v)t)/(2) \\\\3200 = ((6.4\ +\ v)210)/(2) \\\\3200\ *\ 2 = (6.4\ +\ v)210\\\\6400 = 1344\ +\ 210v\\\\6400\ -\ 1344\ = 210v\\\\5056 = 210v\\\\v = (5056)/(210) \\\\v = 24.1\ m/s`