Acceleration and distance travelled questions

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SUMMARY

The discussion focuses on a physics problem involving a car that accelerates from 36 km/h to 78 km/h in 6 seconds and then stops over a distance of 120 meters. The correct rate of acceleration is calculated using the formula for acceleration, which is the change in velocity divided by time. The area under the velocity-time graph represents the distance traveled during acceleration. Additionally, the increase in kinetic energy can be determined by comparing the kinetic energy at different velocities, specifically from 36 km/h to 78 km/h.

PREREQUISITES
  • Understanding of basic kinematics, including acceleration and velocity.
  • Familiarity with the equations of motion relating displacement, velocity, and acceleration.
  • Knowledge of kinetic energy calculations and the formula KE = 0.5 * mass * velocity^2.
  • Ability to interpret and draw velocity-time graphs.
NEXT STEPS
  • Learn how to calculate acceleration using the formula: acceleration = (final velocity - initial velocity) / time.
  • Study the equations of motion to solve for displacement and time in various scenarios.
  • Explore kinetic energy concepts and practice calculating changes in kinetic energy for different velocities.
  • Practice drawing and interpreting velocity-time graphs to visualize motion and calculate distances.
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Students studying physics, educators teaching kinematics, and anyone interested in understanding motion and energy concepts in automotive contexts.

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1. Homework Statement [/b
a car traveling at 36km/h accelerates to 78km/h in 6 seconds. the brakes are applied and the car stops in a distance of 120m, draw a v/t graph and workout

The rate of acceleration? i got 6km/h please correct if this is incorrect
The distance traveled during the acceleration stage?
The time to stop the car?
If the car has a mass of 900kg. determine the increase in kinetic energy?

Homework Equations





The Attempt at a Solution

 
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How about providing some of those equations? If you do that you'll be nearly done solving the problem.

Note that the units of acceleration are displacement/time^2. By how much does the velocity change per second? You accelerate from 36 km/h to 78 km/h in 6 seconds, so you can solve that easily.

If you draw the velocity curve over time, the area below it represents distance traveled, and the slop at any instance represents acceleration.

Else, there are three equations relating displacement, velocity, and acceleration. Those will answer the whole problem except the one about kinetic energy. You can look up that equation also.

"Increase" of KE implies two measurements at different times. Those could be from 36 to 78, from 78 to 0 or from 36 to 0. The latter two cases will yield a negative result.
 
for your solution of acceleration, you are numerically close, but still not quite correct
 

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