How to Calculate Acceleration and Average Speed in a Road Test

In summary, the conversation discusses a problem in which a car is uniformly accelerated from rest over a distance of 400m in 18.5s, then decelerates and stops in 4.9s. The first part of the problem is solved by using the equation v=d/t, resulting in an acceleration of 21.62ms^-1. The conversation then moves on to discussing how to find the average speed of the car for the entire journey, which involves finding the velocity of the deceleration portion. The suggestion is made to split the problem into two parts and use the equation x=ut+0.5at^2 to find the deceleration. A value of 2.34m^-2 is
  • #1
chaos17
4
0

Homework Statement



In a road test,a car was uniformly accelerated from rest over a distance of 400m in 18.5s. The driver then applied the brakes, stopping the car in 4.9 s with constant deceleration.

a) calculate the acceleration of the car for the first 400m.

b)Calculate the average speed of the car for the journey covering both the acceleration and braking distances.

Homework Equations



v=d/t

avg speed=distance / time

The Attempt at a Solution



a)v=d/t

v=400/18.5
=21.62ms^-1

b) I have no idea, I don't know what to do. Could someone give me a hint on what values to use or what formula to use.

Is my part a right?
 
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  • #2
what i mean by part a being right is that do i then use v to work out acceleration? Or is this a wrong way?
 
  • #3
From what I can see, part (a) is right. The first thing that would be good to do is, to split the acceleration and the deceleration sections up. As you know the velocity of the first part, you need to find the velocity of the second part. How would you go about doing that, as you have the time and the initial velocity of the car, but you do not have the deceleration, final velocity or the distance the deceleration acts over. You do not need to know all everything, just enough to find out the average velocity over the whole[b/] time.

In reply to second post by Chaos: I believe you are on the right track. What equation do you use to then work out acceleration, when you have the velocity, time and distance?
 
Last edited:
  • #4
I would them use avg acceleration= change in velocity / time.

Now that I think of it, wouldn't part a be:

u=0, t=18.5, x=400

x=ut+0.5at^2

so 400=0.5 * a * (18.5^2)
a=400/171.125
a=2.34m^-2

I think that is right.

But them what do I do for part b.
 
  • #5
can anyone help for part b?
 

1. What is the difference between motion in 1 dimension and motion in 2 dimensions?

Motion in 1 dimension refers to the movement of an object along a straight line, while motion in 2 dimensions involves movement along two perpendicular axes (x and y). In other words, motion in 1 dimension is restricted to only one direction, while motion in 2 dimensions can occur in any direction.

2. How is velocity different from speed?

Velocity and speed are often used interchangeably, but they are actually distinct concepts. Speed is a measure of how fast an object is moving, while velocity is a measure of both speed and direction of motion. In other words, velocity includes information about the object's speed and the direction in which it is moving.

3. What is acceleration and how is it related to motion in 1 and 2 dimensions?

Acceleration is the rate of change of an object's velocity over time. In motion in 1 dimension, acceleration can only occur in the direction of motion, while in motion in 2 dimensions, acceleration can occur in any direction. In both cases, acceleration can either increase or decrease an object's speed and/or change its direction of motion.

4. How do vectors and scalars relate to motion in 1 and 2 dimensions?

Vectors are quantities that have both magnitude (size or numerical value) and direction, while scalars only have magnitude. In motion in 1 and 2 dimensions, displacement, velocity, and acceleration are all vector quantities, while speed is a scalar quantity.

5. What is the difference between position and displacement?

Position refers to the location of an object, while displacement refers to the change in position. Displacement takes into account both the distance an object has traveled and its direction of motion, while position only indicates where the object is located at a given moment.

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