Question

A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.53 m/s2 and β = 4.80×10−2 m/s3. Calculate the average velocity of the car for each time interval:_______.
a) t = 0 to t = 2.00 s.
b) t = 0 to t = 4.00 s.
c) t = 2.00 s to t = 4.00 s.

Answer 1

Answer:

(a) $V_{avg}=2.868m/s$

(b) $V_{avg}=5.352m/s$

(c) $V_{avg}=7.836m/s$

Explanation:

Given data

x(t)=αt²-βt³

α=1.53m/s²

β=0.0480m/s³

First we need to find distance x at these time so

x(t)=1.53t²-0.0480t³

at t=0

x(0)=1.53(0)²-0.0480(0)³=0m

at t=2

x(2)=1.53(2)²-0.0480(2)³=5.736m

at t=4s

x(4)=1.53(4)²-0.0480(4)³=21.408 m

For(a) Average velocity at t=0s to t=2s

The average velocity is given as

Vavg=Δx/Δt

$V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(2)=5.736m\\V_{avg}=\frac{5.736m-0m }{2s-0s}\\V_{avg}=2.868m/s$

For(b) Average velocity at t=0s to t=4s

The average velocity is given as

Vavg=Δx/Δt

$V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-0m }{4s-0s}\\V_{avg}=5.352m/s$

For(c) Average velocity at t=2s to t=4s

The average velocity is given as

Vavg=Δx/Δt

$V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(2)=5.736m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-5.736m }{4s-2s}\\V_{avg}=7.836m/s$