Calculating Distance to Highway Patrol Car

  • Thread starter Thread starter Naeem
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the distance to a highway patrol car while decelerating from 85 mph to 65 mph at a rate of 4 mph/sec. The first phase of deceleration takes 5 seconds, during which the car reduces its speed. After this, the car maintains a constant speed of 65 mph for an additional 10 seconds before passing the patrol car. The total distance calculation requires using the equations of motion, ensuring consistent units throughout the calculations.

PREREQUISITES
  • Understanding of kinematic equations, specifically x - x0 = v0t + 1/2 at²
  • Knowledge of unit conversion between miles per hour and meters per second
  • Familiarity with concepts of acceleration and deceleration
  • Basic algebra skills for solving equations
NEXT STEPS
  • Learn about unit conversion techniques for speed and acceleration
  • Study the application of kinematic equations in real-world scenarios
  • Explore the implications of constant acceleration in motion problems
  • Investigate how to graphically represent motion using velocity-time graphs
USEFUL FOR

Students in physics, automotive engineers, and anyone interested in understanding motion dynamics and speed calculations in real-life scenarios.

Naeem
Messages
193
Reaction score
0
Q. Suppose you are driving on the interstate at 85 mph when you see a highway patrol car parked up ahead. You apply your brakes, decelerating at a constant rate of 4 mph/sec until you're reduced your speed to 65 mph. You then maintain a constant speed of 65 mph. You pass the patrol car 15 seconds after you initially applied the brakes and are relieved that he doesn't pull out after you. How far away was the highway patrol car when you first saw it?

Ans. Two different situations to this problem,

1st situation,

we can find the time, it took to reduce the speed from 85 mph to 65 mph

t = (85mph - 65mph) / 4 m/s2

t = 5 seconds. ( Note: It decelarates in the first phase @ the rate of

4mph / sec.)

2nd Situation

Thereafter, it continues, at a constant rate of 65 mph.

Use constatnt acc. equation.

x - xo = volt + 1/2 at^2

Will vo be 65mph

and t = 10 seconds

a = -4 mph /sec
-----------------------------------

For the first situation, knowing the time t = 5 seconds,

we need to use

x - xo = volt + 1/2 at^2 again

Here, what would vo be?

t would be 5 seconds for sure.

I think 'a' would be 4mph /sec

Am I right with the variables.

Please help anybody
 
Physics news on Phys.org
Naeem said:
t = (85mph - 65mph) / 4 m/s2

t = 5 seconds. ( Note: It decelarates in the first phase @ the rate of

4mph / sec.)

Your answer looks good, but the units in your equation are off.


x - xo = volt + 1/2 at^2

Will vo be 65mph

and t = 10 seconds

a = -4 mph /sec

It's not clear what you're doing here. If you're solving for the distance the car travels while moving at a constant 65 mph, then "a" should be zero because the velocity is constant. Also, you need consistency with your units. If your times are going to be in seconds, then your velocities and accelerations should be meters (or miles) per second and meters (or miles) per second squared, not miles per hour and miles per hour per second.


For the first situation, knowing the time t = 5 seconds,

we need to use

x - xo = volt + 1/2 at^2 again

Here, what would vo be?

t would be 5 seconds for sure.

I think 'a' would be 4mph /sec

You're right that "t" would be 5 seconds, but "a" would be -4 mph/sec (be sure to convert this before putting it in) and v0 would be 85 mph.
 
Got it !
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
1
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
Replies
2
Views
3K
Replies
9
Views
5K
Replies
45
Views
6K
Replies
1
Views
3K
Replies
5
Views
2K
Replies
4
Views
3K