Question

A car’s initial velocity is 50.0 km/h in the direction 60.0° north of east, and its final velocity is 70.0 km/h in the direction 40.0° south of east. If the time period for this journey is 30.0 minutes, what is the magnitude of the car’s average acceleration? Group of answer choices

186 km/h/h

240 km/h/h

92.8 km/h/h

3.94 km/h/h

Answer 1

_____________________________________________________

Average acceleration is given by:

a_avg = Change in velocity/time taken

_____________________________________________________

time taken 30 min = 0.5 hr

Change in velocity = Vf - Vi

Vi = 50 km/h at 60 deg N of E

Vix = 50*cos 60 deg = 25 km/hr

Viy = 50*sin 60 deg = 43.3 km/hr

Vi = 25 i + 43.3 j

Vf = 70 km/h at 40 deg S of E

Vfx = 70*cos 40 deg = 53.6 km/hr

Vfy = -70*sin 40 deg = -45.0 km/hr

Vf = 53.6 i - 45.0 j

Now

Vf - Vi = (25 i + 43.3 j) - (53.6 i - 45.0 j)

Vf - Vi = (25 - 53.6) i + (43.3 + 45) j

Vf - Vi = -28.6 i + 88.3 j

So now average acceleration will be

a_avg = (Vf - Vi)/(tf - ti)

a_avg = (-28.6 i + 88.3 j)/0.5 hr

a_avg = -57.2 i + 176.6 j

a_avg = -57.2 km/hr^2 i + 176.6 km/hr^2 j

Now Magnitude of a_avg will be

|a_avg| = sqrt ((-57.2)^2 + (176.6)^2) = 185.6 km/hr^2

so correct option is = 186 km/h/h

_____________________________________________________

Please rate
if any mistake in this answer please comment i will clarify your doubt . thank you