Question

a truck is traveling on a straight stretch of of highway at 120 km/h. the driver spots a police car ahead. if the truck slows with an average acceleration of -4.0 m/s2 how much time does it take to reduce its speed to the legal limit of 96 km/h?

Answer 1

Answer:

It takes 1.67 seconds to reduce its speed to the legal limit

Explanation:

The main point in this problem is the difference between the units of

the speeds and acceleration

The unit of the speeds is km/hour

The unit of the acceleration is m/sec²

So we must to change the units of the speeds from km/hour to

meter/second

1 km = 1000 m

1 hour = 60 × 60 = 3600 seconds

Then 120 km/hr = $\frac{120*1000}{3600}=33.33$ m/s

96 km/hr = $\frac{96*1000}{3600}=26.67$ m/s

To find the time of deceleration (to reduced the speed) we will use:

v = u + at. where u is the initial velocity, v is the final velocity, a is the

acceleration and t is the time

v = 26.67 m/s , u = 33.33 m/s , a = - 4 m/s²

Substitute these values in the rule

26.67 = 33.33 + (- 4)t

Subtract 33.33 from both sides

- 6.66 = - 4t

Divide both sides by - 4

t = 1.67 seconds

It takes 1.67 seconds to reduce its speed to the legal limit