How far will a car travel when decelerating from 60km/h to 10km/h in 2.0s?

In summary, the question involves calculating the distance a car will travel when its speed is reduced from 60km/h to 0km/h in 2.0 seconds, assuming the same deceleration as before. The correct approach is to convert all units to meters and seconds, and then use the equation v2^2 = v1^2 + 2(a)(d) to solve for d. After correcting some errors, the final answer is 0.033km.
  • #1
xcompulsion
6
0
A car traveling at 40km/h has it's speed reduced to 10km/h in 2.0s. Assuming that the same deceleration would be in effect, find how far the car will travel in coming to rest from a speed of 60km/h?
 
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  • #2
What are you thoughts on how to start this problem?
 
  • #3
hage567 said:
What are you thoughts on how to start this problem?
I'm thinking about calculating the acceleration using the givens from the first part, to calculate the second part.

so using the equation,
v2^2 = v1^2 + 2(a)(t)

But what has gotten me really puzzled is the seconds, the hours, and the kilometers.
 
  • #4
That's the right approach, but that equation is not quite right. That t should be a d (as in distance). Check for another equation.

You will need to convert your quantities so they are all in the same units. I usually work in meters and seconds in these questions. I would start with that before you do anything else.
 
  • #5
hage567 said:
That's the right approach, but that equation is not quite right. That t should be a d (as in distance). Check for another equation.

You will need to convert your quantities so they are all in the same units. I usually work in meters and seconds in these questions. I would start with that before you do anything else.
Okay, so after converting all the units to m/s..

first part:
v1 = 40km/h = 11.11m/s
v2 = 10km/h = 2.8m/s
t=2.0s
a = v/t = (2.8m/s - 11.11m/s) / 2.0s = -4.155m/s^2

second part:
v1 = 60km/h = 16.67m/s
v2 = 0km/h = 0m/s
a = -4.155m/s^2
d = ?

v2^2 = v1^2 + 2(a)(d)
0 = (16.67)^2 + 2(-4.155)(d)
0 = 277.9 + (-831)(d)
d = 831 - 277.9
d = 553.1m/s

However, the correct answer to this question is 0.033km..
 
  • #6
That -831 is wrong, you missed the decimal place!

Also, you didn't solve for d correctly. There should be division involved. Try again.
 
  • #7
v2^2 = v1^2 + 2(a)(d)
0 = (16.67)^2 + 2(-4.155)(d)
0 = 277.9 + (-8.31)(d)
-277.9 / -8.31 = (-8.31)(d) / (-8.31)
d = 33.44m

so, 33.44/1000 = 0.033km!

thank you so much :smile:
 
  • #8
You're welcome!
 

FAQ: How far will a car travel when decelerating from 60km/h to 10km/h in 2.0s?

What are equations of motion?

Equations of motion are mathematical formulas used to describe the motion of an object in a certain system. They take into account factors such as position, velocity, acceleration, and time to determine the movement of an object.

How are equations of motion used in science?

Equations of motion are used in various fields of science, including physics, engineering, and astronomy. They help us understand and predict the behavior of objects and their motion in different systems.

What are the three basic equations of motion?

The three basic equations of motion are:
1. Velocity equation: v = u + at
2. Displacement equation: s = ut + 1/2at²
3. Acceleration equation: v² = u² + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement

Can equations of motion be used for both linear and circular motion?

Yes, equations of motion can be applied to both linear and circular motion. However, for circular motion, additional equations such as centripetal acceleration and centripetal force are also needed to fully describe the motion.

Are there any limitations to using equations of motion?

Equations of motion are based on certain assumptions and may not accurately describe the motion of objects in all situations. They do not take into account factors such as air resistance, friction, and other external forces that may affect the motion of an object. Therefore, they may have limitations in real-life scenarios.

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