A car is travelling at a speed of 13 m/s. The driver applies the brakes and the car comes to rest after travelling a further 26 m. What is the deceleration of the car in m/s^2?
Vi = initial speed
v = Vi + a t
easy way to get time is use average speed during constant acceleration
Vaverage = Va = (13+0) /2 = 6.5 m/s
so time to stop = t = 26 / 6.5 = 4 seconds
so
0 = 13 + a (4)
a = -13/4 m/s^2
To find the deceleration of the car, we will use the following equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (13 m/s)
s = displacement (26 m)
Rearranging the equation to solve for the acceleration (a), we have:
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (0^2 - 13^2) / (2 * 26)
a = (-169) / 52
a ≈ -3.25 m/s^2
Therefore, the deceleration of the car is approximately -3.25 m/s^2.
To find the deceleration of the car, we can use the formula:
deceleration = change in velocity / time
First, let's find the change in velocity. The car comes to rest, so its final velocity is 0 m/s. The initial velocity is given as 13 m/s. Therefore, the change in velocity is:
change in velocity = 0 - 13 = -13 m/s
Next, we need to find the time it takes for the car to come to rest. We can use the equation of motion:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
Since the final velocity is 0 and the initial velocity is 13 m/s, we can simplify the equation to:
0 = 13^2 + 2 * acceleration * 26
169 = 52 * acceleration
acceleration = 169 / 52
acceleration ≈ 3.25 m/s^2
Therefore, the deceleration of the car is approximately 3.25 m/s^2.