A race car moves such that its position fits the relationship x = (5.5 m/s) t +
ID: 1653594 • Letter: A
Question
A race car moves such that its position fits the relationship
x = (5.5 m/s)t + (0.85 m/s3)t3
where x is measured in meters and t in seconds.
(a) A plot of the car's position versus time is which of the following?
(b) Determine the instantaneous velocity of the car at t = 4.6 s,using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to better see the limiting process keep at least three decimal places in your answer.)
t = 0.40 s
t = 0.20 s
t = 4.50 s
t = 0.10 s
m/s (Use the interval from
t = 4.55 s
Explanation / Answer
Given
position of the car as x = 5.5 m/s*t+0.85 m/s3 *t^3
x = 5.5*t+0.85*t^3 m
if we substitute the value of t = 1s , the position is x = 5.5*1+0.85*1^3 m = 6.35 m
and substitute the value of t = 2 s , the position is x = 5.5*2+0.85*2^3 m = 17.8 m
substitute the value of t = 3 s , the position is x = 5.5*3+0.85*3^3 m = 39.45 m
substitute the value of t = 4 s , the position is x = 5.5*4+0.85*4^3 m = 76.4 m
these values will be fit in the second graph only.
b)
instantaneous velocity is dx/dt = 5.5+3*0.85t^2
at time t = 4.6 s is
v(4.6) = 5.5+3*0.85*4.6^2 m/s = 59.458 m/s
for Dt = 0.4 s at times t1= 4.4s, t2 = 4.8 s is
v = (x2-x1)/(t2-t1) m/s
v = (x(4.8)-x(4.4)/(4.8-4.4)
v = (120.4-96.6)/0.4 m/s
V = 59.5 m/s
for Dt = 0.2 s is at times t2= 4.7s , t1 = 4.5 s is
v = (x2-x1)/(t2-t1) m/s
v = (x(4.7)-x(4.5)/(4.7-4.5)
v = (114.1-102.2)/0.2 m/s
V = 59.5 m/s
for Dt = 0.1 s is at times t2= 4.55 s , t1 = 4.65 s is
v = (x2-x1)/(t2-t1) m/s
v = (x(4.65)-x(4.55)/(4.65-4.55)
v = (111.03-105.09)/0.1 m/s
V = 59.4 m/s