Question

A car starts from rest on a curved road of 250 m radius and accelerates at a constant tangential acceleration of 0.6 m/s^2. Determine the distance and time for which that car will travel before the magnitude of total acceleration attained by it becomes 0.75m/s^2.​

Answer 1

Answer:

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Answer 2

Answer:

The distance and the time for the car are 124.84m and 23.5sec respectively.

Explanation:

The total acceleration of the car is given as,

$a=\sqrt{a^2_t+a^2_n}$      (1)

Where,

a=total acceleration of the car

at=tangential acceleration of the car

an=normal acceleration of the car

From the question we have,

at=0.6m/s²

a=0.75m/s²

The radius of the curved road(r)=250m

By substituting the value of at and a in equation (1) we get;

$0.75=\sqrt{(0.6)^2+a^2_n}$

$a_n=\sqrt{(0.75)^2-(0.6)^2} =\sqrt{0.2025}=0.45m/s^2$

Then,

$a_n=\frac{v^2}{r}$      (2)

$v=\sqrt{a_n r} =\sqrt{0.45\times 250} =10.60m/s$

From the first equation of motion we have,

$v=u+ta_n$      (3)

$10.60=0+t\times 0.45$  (u=0 as the car starts from rest)

$t=\frac{10.60}{0.45} =23.5sec$  

From the third equation of motion we have,

$v^2-u^2=2a_ns$

$(10.60)^2-(0)^2=2\times 0.45\times s$

$s=\frac{(10.60)^2}{2\times 0.45} =124.84m$

Hence, the distance and the time for which that car will travel before the magnitude of total acceleration attained by it becomes 0.75m/s² is  124.84m and 23.5sec respectively.