Question

The car travels along the circular path such that its speed is increased by at = (0.5et ) m/s2 , where t is in seconds. Determine the magnitudes of its velocity and acceleration after the car has traveled s =18 m starting from rest. Neglect the size of the carDetermine the magnitude of its acceleration after the car has traveled s = 18 m starting from rest

Answer 1

Answer:

acceleration = 24.23 ms⁻¹

Explanation:

Let's gather the data:

The acceleration of the car is given by a = 0.5 $e^{t}$

The acceleration is the change in the speed in relation to time. In other words:

$\frac{dv}{dt}$ = a = a = 0.5 $e^{t}$            ...1

Solving the differential equation yields:

v = 0.5 $e^{t}$ + C₁

Initial conditions : 0 = 0.5 $e^{0}$ + C₁

                             C₁ = -5

at any time t, the velocity is: v= 0.5$e^{t}$- 5

Solving for distance, s = 0. 5$e^{t}$ - 0.5 t - 0.5

                                    18 = 0.5 $e^{t}$ - 0.5 t - 0.5

                                       t = 3.71 s

Substitute t = 3.71 s

                   v= 0.5$e^{t}$- 5

                     = 19.85 m/s

a = 0.5 $e^{t}$             ...1

 = 20.3531

an = $\frac{v^{2} }{p}$

     = $\frac{(19.853)^{2} }{30}$

     = 13.1382

Magnitude of acceleration = $\sqrt{ (a)^{2} + (an)^{2} }$

                                            = $\sqrt{(20.3531)^{2} +(13.1382)^{2} }$

                                            = 24.23 ms⁻¹ Ans