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?
A)186 km/h/h
B)240 km/h/h
C)92.8 km/h/h
D)3.94 km/h/h

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?​
A)186​ km/h/h
B)​240 km/h/h
C)​92.8 km/h/h
D)​3.94 km/h/h

Quizplus · Accepted Answer

Explanation:The magnitude of the average acceleration is given by:$$a_{avg}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t}$$where:* $v_i$ is the initial velocity* $v_f$ is the final velocity* $t$ is the time periodIn this problem, we have:* $v_i=50.0	ext{ km/h}$ at $60.0\degree$ north of east* $v_f=70.0	ext{ km/h}$ at $40.0\degree$ south of east* $t=30.0	ext{ minutes}=0.5	ext{ hours}$We can first convert the velocities to their x and y components:$$v_{ix}=v_i\cos	heta_i=(50.0	ext{ km/h})\cos60.0\degree=25.0	ext{ km/h}$$$$v_{iy}=v_i\sin	heta_i=(50.0	ext{ km/h})\sin60.0\degree=43.3	ext{ km/h}$$$$v_{fx}=v_f\cos	heta_f=(70.0	ext{ km/h})\cos40.0\degree=53.2	ext{ km/h}$$$$v_{fy}=v_f\sin	heta_f=(70.0	ext{ km/h})\sin40.0\degree=44.3	ext{ km/h}$$The change in velocity is then:$$\Delta v=\sqrt{(v_{fx}-v_{ix})^2+(v_{fy}-v_{iy})^2}=\sqrt{(53.2	ext{ km/h}-25.0	ext{ km/h})^2+(44.3	ext{ km/h}-43.3	ext{ km/h})^2}=32.8	ext{ km/h}$$The average acceleration is then:$$a_{avg}=\frac{\Delta v}{\Delta t}=\frac{32.8	ext{ km/h}}{0.5	ext{ hours}}=\boxed{65.6	ext{ km/h/h}}$$Therefore, the magnitude of the car's average acceleration is 65.6 km/h/h, which is choice A.

A Car's Initial Velocity Is 50