# Acceleration and velocity

roam

## Homework Statement

A racehorse accelerates from rest at time t = 0.0 s, out of its starting stalls with a time dependent acceleration a given by a = 1.7α - 0.6β t. It accelerates so until time = 4 s when it reaches a consant racing speed of $$v = 15.5 ms^{–1}$$.

(a) Use this information to first estabilsh how the constants α and β are related to each other.

(b) Using the definition of acceleration as the time rate of change of velocity, find an expression for the velocity and use it to establish the values of the consants α and β from the information available above. Be sure to include the correct SI units with the values of each of these constants.

## The Attempt at a Solution

(a) The answer has to be $$\alpha = 1.41 \beta$$ and when I solve the following for alpha in terms of beta

$$a =1.7 \alpha - 0.6 \beta 4$$

$$a=v/t = 15.5/4=3.87$$ so

$$3.9= 1.7 \alpha -2.4 \beta$$

$$\alpha = 2.29+1.41 \beta$$

So it should be $$\alpha = 2.29+1.41 \beta$$, but why does the answers suggest that is should be $$\alpha = 1.41 \beta$$? What happens to the term 2.29?

(b) It wouldn't work if I substitute $$\alpha= 1.41 \beta$$ into that formual like $$1.41 \beta = 2.29+1.41 \beta$$. So how do I use $$\alpha = 2.29+1.41 \beta$$ (definition of acceleration as the time rate of change of velocity) to find the vales of alpha & beta?

Btw, the answer should be $$\alpha = 4.56 ms^{-2}, \beta=3.23 ms^{-3}$$.

## Answers and Replies

Homework Helper
at t=4, the speed is constant and so is the direction of travel, so the velocity is constant.

What does this mean for the acceleration at t=4?

a(t)= 1.7α - 0.6β t, a(4) = ?

roam
at t=4, the speed is constant and so is the direction of travel, so the velocity is constant.

What does this mean for the acceleration at t=4?

a(t)= 1.7α - 0.6β t, a(4) = ?

oh, that works! Thanks!!

I also don't get part (b). I use the formula $$a = \frac{v}{t} \Rightarrow v=at$$ and since $$a =1.7 \alpha - 0.6 \beta t$$ we have

$$15.5=(1.7 \alpha - 0.6 \beta t)t$$

$$15.5 = (1.41 \beta - 0.6 \beta t)t = 1.41 \beta t-0.6 \beta t^2$$

At t=4

$$15.5 = 5.64 \beta -9.6 \beta$$

$$\beta = -3.96$$

I don't know why my answer for beta is different from the correct answer ($$3.23 ms^{-3}$$). My answer is negative. By the way, why is it that the unit for beta is $$ms^{-3}$$?

Homework Helper
if a=dv/dt, then v=∫a dt.

Find ∫a dt to get v(t)

roam
if a=dv/dt, then v=∫a dt.

Find ∫a dt to get v(t)

Are you sure? Because this doesn't seem to work. Here I tried it:

$$a=1.7 \alpha - 0.6 \beta t$$

$$\int^{4}_{0} 1.7 \alpha - 0.6 \beta t = \int^{4}_{0} 1.41 \beta t - 0.6 \beta t$$

$$= 1.41 \beta t - \frac{0.6 \beta t^2}{2}|^4_0 = 5.64 \beta - 4.8 \frac{\beta}{2}=0$$

$$\frac{6.48 \beta}{2} \Rightarrow \beta = 0.30$$

This is still wrong because the answer must be 3.23.

Homework Helper
∫1.7α dt = 1.7αt

roam
∫1.7α dt = 1.7αt

I'm not quite sure what I'm supposed to be doing...now we have $$1.7 \alpha t-(0.6 \beta)\2 t^2$$, for t=4 it is $$6.8 \alpa -4.8 \beta$$. But if I substitute $$\alpha = 1.41 \beta$$ into it, the expression becomes $$1.7 \alpha t-(0.6 \beta)\2 t^2 =9.58 \beta - 4.8 \beta = 4.78 \beta$$. But this doesn't tell us anything about the value of beta! am I missing something?

Homework Helper
so you have $v=1.7 \alpha t -(0.6 \beta)\frac{t^2}{2}$

at t=4, v=15.5

roam
Thanks I figured it out.

Last question: determine how far this horse travels during this 4 s start up phase of the race. (correct answer is 41.3 m)

What's the best equation to use here? I know that $$x=vt, 15.5 \times 4 = 62 \neq 41.3$$.

Homework Helper
a = dv/dt = 1.7α - 0.6β*t
dv = (1.7α - 0.6β*t)*dt
v = 1.7α*t - 0.6β*t^2/2
dx = (1.7α*t - 0.6β*t^2/2)*dt
So x = (1.7α*t^2)/2 - (0.6β*t^3)/2*3
Substitute the values of α and β, and find x.