please try to solve all of them Thanks! a previous | 1 of 9 next a Part C Exerci

Question

please try to solve all of them

Thanks!

a previous | 1 of 9 next a Part C Exercise 3.8 What is the magnitude of the velocity of the car at t = 6.34 s ? Express your answer to three significant figures and include the appropriate units. A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by ms- 2.00 m/s(0.550 m/s2)tj Value Units Submit My Answers Give Up Part D What is the direction (in degrees counterclockwise from +x-axis) of the velocity of the car att-6.34 s? Express your answer to three significant figures and include the appropriate units. 8, 117.34 Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Part E

Explanation / Answer

Given

velocity of the car as a function of time is

V = (5.00 m/s -(0.0180 m/s^3)t^2) i +(2 m/s+(0.550 m/s2)t) j

V = (5.00 -(0.0180 )t^2) i +(2 +(0.550 )t) j

Part C

magnitude of the velocity of the car at t = 6.34 s is

substitting t = 6.34 s in the above eq

v(6.34) = (5.00 -(0.0180 )6.34^2) i +(2 +(0.550 )6.34) j

= 4.2764792 i + 5.487 j

magnitude is V = sqrt((4.2764792)^2+(5.487)^2) m/s = 6.9567 m/s

Part E

direction is tan theta = (Vy/vx) = theta = arc tan (vy/vx) = arc tan (5.487/4.2764792 ) degrees = 52 degrees

Part F

accelerationof the car at t = 6.34 s is

differentiating v(t) w.r.t time once we get

a(t) = (-2*(0.0180 m/s^3)t) i +((0.550 m/s2)) j

a(6.34) = (-2*(0.0180 )6.34) i +((0.550 )) j

a(6.34) = -0.22824 i + 0.550 j

magnitude is a = sqrt((-0.22824)^2+(0.550)^2) m/s2 = 0.5955 m/s2

Part F

the direction of the acceleration at t = 6.34s is

theta = arc tan (ay/ax)

= arc tan (0.550/(-0.22824)) degrees

= -67.4624 degrees that is from x axis the angle is 360-67.4624 = 292.5376 degrees

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