Question

A driver in a moving car applies the brakes. The car slows to a final speed of 1.50 m/s over a distance of 40.0 m and a time interval of 9.00 s. The acceleration while braking is approximately constant. (a) What is the car's original speed before braking? m/s (b) What is its acceleration during this time? (The car's initial velocity is in the positive direction. Indicate the direction with the sign of your answer.) m/s2

Answer 1

Solution:

Let us go to the basics first.

Given:

v = 1.50 m/s

S = 40.0 m

t = 9.00 s

a) What is the car's original speed before braking?

From Newton's equation of motion, we know that:

v = u + at [ u = initial/original speed; a = acceleration]

=>1.50 = u + 9a ...................Eqn.1

Also, From Newton's eqn:

v² = u² + 2as

=>2.25 = u² + 80a.................Eqn.2

Now, multiply eqn.1 by (80/9) on both sides:

1.50*(80/9) = (u + 9a)*(80/9)

=>13.333 = (80/9)u + 80a ..............Eqn.1.1

Perform, Eqn.2 minus Eqn 1.1:

2.25 - 13.333 = u² + 80a - (80/9)u - 80a

=>-11.083 = u² - (80/9)u

=> 0 = u² - (80/9)u + 11.083

=>u² - 8.889u + 11.083 =  0

Solving this quadratic eqn., we get:

u = 7.389 or 1.5

Discarding u = 1.5 (as 1.5 m/s is the final velocity given in the question)

Thus, Original speed = u = 7.389 m/s (Answer a)

(b) What is its acceleration during this time?

From eqn.1, we have:

1.50 = u + 9a

=> 1.50 = 7.389 + 9a

=> a = -0.6543 m/s² (Answer b)

Direction: Opposite to the motion of car (i.e., in negative direction)

Thanks!!!