Respuesta :
(a). The car's average velocity between t = 1.0s to t = 1.5s will be - [tex]1\;m/s[/tex]
(b). The car's acceleration at t = 1.5s will be - [tex]0.4\;m/s^{2}[/tex]
(c). Car's speed is increasing with time.
We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.
We have to determine -
- The car's average velocity (in m/s) in the interval between -
t = 1.0 s to t = 1.5 s.
- The car's acceleration at t = 1.5 s.
- Determining whether car's speed increasing or decreasing with time.
What is Acceleration?
The rate of change of velocity with respect to time is called Acceleration. Mathematically -
[tex]$a=\frac{dv}{dt}[/tex]
According to the question, we have the following data for the Car -
t = 0s → x = 0m
t = 0.5s → x = 0.1m
t = 1.0s → x = 0.4m
t = 1.5s → x = 0.9m
t = 2.0s → x = 1.6m
PART - A
The car's average velocity between t = 1.0s to t = 1.5s will be -
[tex]$v_{avg} = \frac{0.9-0.4}{1.5-1}= 1 m/s[/tex]
PART - B
Velocity at t = 1.5 s will be -
[tex]$v(1.5)=\frac{0.9}{1.5}= 0.6\;m/s[/tex]
The car's acceleration at t = 1.5s will be -
[tex]$a(1.5) = \frac{v}{t} = \frac{0.6}{1.5} = 0.4\;m/s^{2}[/tex]
PART - C
Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.
[ The following data was missing in your answer. The complete question would include this data also -
t = 0s → x = 0m
t = 0.5s → x = 0.1m
t = 1.0s → x = 0.4m
t = 1.5s → x = 0.9m
t = 2.0s → x = 1.6m ]
To solve more questions on Kinematics, visit the link below-
https://brainly.com/question/17272824
#SPJ2