Question

A car accelerates from rest at a constant rate a(alpha) for some time after which it decelerates at a constant rate B(beta) to come to rest. If the total time is t prove
that maximum velocity reached is
Vm = aBt/a + B
and total displacement is
s = aBt^2/2(a+ß)​

Answer 1

Answer:

If the car accelerates for time $t_1$ and declerates for time $t_2$

Then, the maximum velocity will be when it stops accelerating

Using the first equation of motion

$V_m=0+\alpha\times t_1$

$\implies t_1=\frac{V_m}{\alpha}$

Also,

$0=V_m-\beta \times t_2$

$\implies t_2=\frac{V_m}{\beta}$

Given $t_1+t_2=t$

Therefore,

$\frac{V_m}{\alpha}+\frac{V_m}{\beta}=t$

$\implies \frac{V_m(\alpha+\beta)}{\alpha\beta}=t$

$\implies V_m(\alpha+\beta)=\alpha\beta t$

$\implies V_m=\frac{\alpha\beta t}{\alpha+\beta}$

Again, using the third equation of motion

$V_m^2=0^2+2\alpha s_1$

$\implies s_1=\frac{V_m^2}{2\alpha}$

Similarly,

$s_2=\frac{V_m^2}{2\beta}$

Total displacement

$s=s_1+s_2$

$\implies s=\frac{V_m^2}{2\alpha}+\frac{V_m^2}{2\beta}$

$\implies s=V_m^2\times \frac{\alpha+\beta}{2(\alpha\beta)}$

$\implies s=(\frac{\alpha\beta t}{\alpha+\beta})^2\times \frac{\alpha+\beta}{2(\alpha\beta)}$

$\implies s=\frac{\alpha\beta t^2}{2(\alpha+\beta)}$

Hope this answer was helpful.

Answer 2

Explanation:

here is ur answer dear to me and

.this is direct method