Jason while driving on Kukum highway at 70 m.s – 1 seeing the traffic light turn red, he applies the brake and comes to rest in a time of 2 seconds. What is his deceleration?
You can use the following equation of motion: v = u + at
v is 0, since the vehicle comes to a stop.
u is given to be 70m/s
t is 2 seconds
Solve for a. You will get a negative value, which implies deceleration.
To find the deceleration, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity, u = 70 m/s
Final velocity, v = 0 m/s
Time taken, t = 2 seconds
Using the formula, we can substitute the values:
acceleration = (v - u) / t
= (0 - 70) / 2
= -70 / 2
= -35 m/s^2
Therefore, Jason's deceleration is -35 m/s^2.
To find Jason's deceleration, we can use the equation:
acceleration = (final velocity - initial velocity) / time
Here, the initial velocity (v0) is 70 m/s (the speed at which Jason was driving), the final velocity (v) is 0 m/s (since he comes to rest), and the time (t) is 2 seconds.
Substituting these values into the equation, we get:
deceleration = (0 - 70) / 2
Simplifying the equation, we have:
deceleration = -70 / 2
Solving the equation, we get:
deceleration = -35 m/s²
Note: The negative sign indicates that Jason is decelerating (slowing down) since his velocity is decreasing.