Question

A motorist traveling on a curved section of highway of radius 2500 ft at the speed of 60 mph. The motorist suddenly applies brakes, causing the automobile to slow down at a constant rate. Knowing that after 8 seconds the speed has been reduced to 45 mph, determine the acceleration of the automobile immediately after the brakes have been applied.

Answer 1

Step-by-step explanation:

2ft/s2

4.14ft/s2

3.10ft/s2

2.75ft2/s

Solution :

(b) Tangential acceleration

=at=

vt-vi

t

=(

45-60

8

)(

22

15

)=-

11

4

ft/s2

The radial acceleration is

an=

v2

r

=

(60×

22

15

)2

2500

=3.1

Answer 2

The acceleration of the automobile is 4.14ft/ s^2.

Given:

r = 2500ft

v1 = 60 mile/h

v2 = 45 mile/h

t = 8 sec

To Find:

The acceleration of the automobile

Solution:

As we know that,

Tangential acceleration of automobile is

$\sf a_t = \dfrac{v_2 - v_1}{t}$

.

$\: \: \sf \: \frac{45 - 60}{8} \\ \\ \: \: \sf \: = - \frac{15}{8} \frac{mile}{s} \\ \\ \: \: \sf \: - \frac{15}{8} \times \frac{22}{15} \\ \\ \: \: \sf \: - \frac{11}{4} \frac{ft}{ {sec}^{2} }$

The radial accleration of automobile is

$\sf a_r = \dfrac{{v}^{2}}{r}$

$\: \: \sf \: \frac{ {(60 \times \frac{22}{15}) }^{2}}{2500} \\ \\ \: \: \: \sf \: = \frac{ {88}^{2} }{2500} \\ \\ \: \: \: \: \sf \: = 3.09 \frac{ft}{ {sec}^{2} }$

The acceleration of automobile is

$\: \: \sf \: a = \sqrt{ {(at)}^{2} + {(ar)}^{2} } \\ \\ \: \: \: \sf \: = \sqrt{ {( - \frac{ 11}{4}) }^{2} + {(3.09)}^{2} } \\ \\ \: \: \: \sf \: = \sqrt{7.56 + 9.54} \\ \\ \: \: \sf = \sqrt{17.1 } \\ \\ \: \: \: \sf \: = 4.135 \\ \\ \: \: \: \sf \: = 4.14 \frac{ft}{ {sec}^{2} }$

______________________________