Question

A motorist is traveling on a curved portion of highway of radius 350 m at a speed of 72 km/h. The brakes are suddenly applied, causing the speed to decrease at a constant rate of 1.25 m/s2. Determine the magnitude of the total acceleration of the automobile (a) immediately after the brakes have been applied, (b) 4 s later.

Answer 1

Explanation:

a1= -1.25m/s²

v = 72km/hr

= 20 m/s

radius=350 m

an = v² / p

normal acceleration = 20²/350

= 1.143 m/s²

total acceleration = root of a1²+a2²

=1.693 m/s²

or 1.7 m/s²......

hope it helps u my friend

Answer 2

Explanation:

Given that,

Radius of the circular path, r = 350 m

Speed of the motorist, v = 72 km/h = 20 m/s

Tangential acceleration, $a_t=-1.25\ m/s^2$

(a) The centripetal acceleration of the motorist is :

$a_c=\dfrac{v^2}{r}\\\\a_c=\dfrac{(20)^2}{350}\\\\a_c=1.14\ m/s^2$

The magnitude of the total acceleration of the automobile is :

$a=\sqrt{a_t^2+a_c^2} \\\\a=\sqrt{(-1.25)^2+(1.14)^2} \\\\a=1.69\ m/s^2$

(b) At t = 4s, velocity,

$v=a_tt\\\\v=1.25\times 4\\\\v=5\ m/s$

Now centripetal acceleration,

$a_c=\dfrac{v^2}{r}\\\\a_c=\dfrac{(5)^2}{350}\\\\a_c=0.0714\ m/s^2$

Total acceleration,

$a=\sqrt{a_t^2+a_c^2} \\\\a=\sqrt{(-1.25)^2+(0.0714)^2} \\\\a=1.25\ m/s^2$

Learn more,

Kinematics

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