# Physics 11 Speed and acceleraton

• anOldMan
In summary: However, when I tried to use the same equation for the case of constant acceleration, I realized I made a mistake. In any case, it will take me a while to wrap my mind around these questions.
anOldMan
[SOLVED] Physics 11 Speed and acceleraton

## Homework Statement

Linda sees an elephant dart into the road 50.0m ahead of her car while she is driving at 65km/hr. She slams on her brakes, which decelerate the car at the rate of -5.80m/s^2. Will she be able to avoid hitting the elephant? Find her stopping distance.

Answer is Yes, stopping distance is 28.1

## The Attempt at a Solution

65km/hr = 18.056m/s $$V_{avg}9.028$$
a = v / t -5.80m/s^2 = 9.028m/s / T = 1.56 seconds
using $$d = v_{0}t+.5at^{2}$$
d = .5(-5.80m/s^2) (2.43s^2) = 7meters
I am missing something here.

Thanks for the help on the previous question and thanks for any help for this question. Your quick response is appreciated.

I seem to have found the answer, but why is it that the rate is not averaged? The rate is going down from 65 to 0. Therefore, the rate should be an average before doing the calculations. I am confused.

V=0
U=18M/S
A=-5.80 M/S^2

V^2=U^2+2AS
>0=324-11.6S
>S=27m

why would you average the rate?since the deceleration is uniform..u must not average
the rate.this shall lead to loss in generality of the problem.

physixguru said:
V=0
U=18M/S
A=-5.80 M/S^2

V^2=U^2+2AS
>0=324-11.6S
>S=27m
Careful with rounding errors.

anOldMan said:
I seem to have found the answer, but why is it that the rate is not averaged? The rate is going down from 65 to 0. Therefore, the rate should be an average before doing the calculations. I am confused.

By rate, do you mean acceleration? For the case of constant acceleration, average acceleration = instantaneous acceleration.

Thanks guys.

why would you average the rate?since the deceleration is uniform..u must not average
the rate.this shall lead to loss in generality of the problem.

This solved my question. I thought about it again and realized that I made an error. In any case, it will take me a while to wrap my mind around these questions. The deceleration is uniform and will go from 65 km/h to 0 km/h in a certain amount of time based on the rate of deceleration. The reason I was averaging the number was because of the formula sheet. It is written as
note: (change) = a triangle. I assume the triangle in front of the variable means change.
a=(change)v/(change)t

I assumed the change meant average.

## 1. What is the difference between speed and acceleration?

Speed is the measure of how fast an object is moving, while acceleration is the measure of how quickly an object's speed is changing. Speed is a scalar quantity, meaning it has only magnitude, while acceleration is a vector quantity, meaning it has both magnitude and direction.

## 2. What is the formula for calculating average speed?

The formula for average speed is total distance traveled divided by total time taken. Mathematically, it can be represented as: average speed = total distance / total time.

## 3. How does velocity differ from speed?

Velocity is a measure of an object's speed and direction of motion. It is a vector quantity, meaning it has both magnitude and direction, while speed is a scalar quantity, having only magnitude. In other words, velocity not only tells us how fast an object is moving, but also the direction in which it is moving.

## 4. Can an object have a constant speed and changing acceleration?

Yes, an object can have a constant speed and changing acceleration. This can occur when the speed remains the same, but the direction of motion changes, causing the object to accelerate.

## 5. What is the relationship between acceleration and force?

Acceleration is directly proportional to the net force acting on an object. This means that as the net force increases, the acceleration of the object also increases. This relationship is described by the formula: acceleration = net force / mass.

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