Answer to Question #162708 in Mechanical Engineering for loy

Question

A race car enters the circular portion of a track that has a radius of 70 m. When the car enters the curve at point P, it is traveling with a speed of 120 km/h that is increasing at 4 m/s2 . Four seconds later, determine the x and y components of velocity and acceleration of the car.

Answer 1

Radius of car= 70m,travelling with speed=120 km/h= 120$\times \frac{5}{18}=33.33\frac{m}{s}$ ,

acceleration=4m/s2, time=t=4s,

we take inital position of particles as

$x=rcos\theta {i} +rsin\theta {j},$

$v=-r\omega sin\theta{i}+r\omega cos\theta{j}$

and

acceleration

$a=-r(\omega^2 cos\theta+\alpha sin\theta){i}+r(\omega^2sin\theta+\alpha cos\theta){j}$

Now, we know that initially angle $\theta = 0^o$

$v_i=$ $r\omega {i}$ , $\omega_i=\frac{33.33}{70}=0.476 rad/s$

Now, after time t= 4 ,

$\omega_f= \frac{16+33.33}{70}=0.7047 rad/s$

$\alpha=\frac{\omega_f-\omega_i}{4}=0.057 \frac{rad}{s^2}$

this is angular acceleration of car

$\theta=\omega_i t+0.5\alpha t^2=0.476\times 4+0.5\times 0.057 \times 16=135.286 ^o$

this is the angular dispalcement of car after 4 second

Now,

$V_f=-70\times .7047 sin(135.286){i}+70\times 0.7047 cos (135.286){j}$

V$_f$ =$-34.706 {i} -35.054{j}$

So here x component of velocity will be $v_x=$ -34.706 m/s and y component of velocity is $v_y=$ 35.054 m/s

Now acceleration of particles will be

$a=-70(0.7047^2 cos(135.285)+ 0.057sin(135.286){i}+70(0.7047^2sin(135.286)+0.057cos (135.286)){j}$

a=-21.89 i- 21.89j

a=30.96 m/s2