A car with an initial velocity of 8 m/s experiences a constant acceleration for 3 sec, after which the velocity is 17 m/s.
a) What was the acceleration?
b) How far did the car travel during that time interval?
What would the equations be for each of these questions?
(17 -8)/3 9/3 = 3 m/sec/sec
a) (initial vel - final vel)/time
b) distance = avg vel x time
To solve these questions, you need to use the equations of motion, specifically the equations of linear motion with constant acceleration. The two equations you will need are:
1) v = u + at
This equation relates the final velocity (v) of an object to its initial velocity (u), acceleration (a), and time (t).
2) s = ut + (1/2)at^2
This equation relates the displacement (s) of an object to its initial velocity (u), time (t), and acceleration (a).
Now, let's address each question separately:
a) What was the acceleration?
Using the first equation, we can rearrange it to solve for acceleration (a):
a = (v - u) / t
Given:
Initial velocity (u) = 8 m/s
Final velocity (v) = 17 m/s
Time (t) = 3 seconds
Substitute these values into the equation:
a = (17 - 8) / 3
a = 9 / 3
a = 3 m/s²
So, the acceleration of the car is 3 m/s².
b) How far did the car travel during that time interval?
Using the second equation, we can rearrange it to solve for displacement (s):
s = ut + (1/2)at²
Given:
Initial velocity (u) = 8 m/s
Time (t) = 3 seconds
Acceleration (a) = 3 m/s²
Substitute these values into the equation:
s = (8 × 3) + (1/2) × 3 × (3)²
s = 24 + (1/2) × 3 × 9
s = 24 + 13.5
s = 37.5 meters
Therefore, the car traveled a distance of 37.5 meters during that time interval.