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?
Select one:
a. -17.5 m.s – 2
b. -35 m.s – 2
c. 17.5 m.s – 2
d. 35 m.s – 2
17.5 m/s2
To find Jason's deceleration, we can use the equation of motion:
v = u + at
Where:
v = final velocity (0 m/s, as he comes to rest)
u = initial velocity (70 m/s)
a = acceleration or deceleration (what we need to find)
t = time taken (2 seconds)
Rearranging the equation, we get:
a = (v - u) / t
Substituting the values, we have:
a = (0 - 70) / 2
a = -70 / 2
a = -35 m/s²
Therefore, Jason's deceleration is -35 m/s².
The correct answer is (b) -35 m/s².
To find Jason's deceleration, we need to use the equation for deceleration:
Deceleration (a) = Change in velocity (Δv) / Time taken (t)
In this case, Jason is driving at a speed of 70 m/s and comes to rest in 2 seconds. Since he comes to rest, his final velocity (vf) will be 0 m/s.
Δv = vf - vi = 0 - 70 = -70 m/s (note the negative sign indicates deceleration)
Now we can calculate the deceleration:
a = Δv / t = -70 m/s / 2 s = -35 m/s²
Therefore, the correct answer is:
b. -35 m.s⁻²
v = Vi + a t
v = 0 at stop
a = -Vi/t = -35 m/s^2