Driving in a car with a constant velocity of 11.4 m/s you suddenly see an obstacle ahead. Immediately applying the brakes the car slows with an acceleration of 3.80 m/s2.
(i) Calculate the stopping distance of the car, and
(ii) the time taken to stop.
Do step (ii) first.
(ii) Vo = a*t
t = 11.4/3.80 seconds
(i) Stopping distance = (avg. velocity)*t
= (Vo/2)*(Vo/a) = Vo^2/(2 a)
You could do question (i) first using
Vo^2 = 2 a X
but I find it more difficult to remember that formula.
thanks alot :D i got 3s for t for part 2!! nd i got 17.1s for part1 is dat correct??
Yes!
post it.
To calculate the stopping distance of the car and the time taken to stop, we can use the equations of motion.
(i) Stopping Distance:
The stopping distance of the car can be determined using the following equation:
V^2 = V0^2 + 2 * a * d
Where:
V = final velocity (0 m/s, as the car comes to a stop)
V0 = initial velocity (11.4 m/s)
a = acceleration (-3.80 m/s^2, as it's deceleration in this case)
d = stopping distance (unknown)
Rearranging the equation, we have:
d = (V^2 - V0^2) / (2 * a)
Plugging in the values, we get:
d = (0 - 11.4^2) / (2 * -3.80)
d = (-129.96) / (-7.60)
d = 17.09 meters
Therefore, the stopping distance of the car is 17.09 meters.
(ii) Time taken to stop:
We can calculate the time taken to stop using the equation:
V = V0 + a * t
Where:
V = final velocity (0 m/s)
V0 = initial velocity (11.4 m/s)
a = acceleration (-3.80 m/s^2)
t = time taken to stop (unknown)
Rearranging the equation, we have:
t = (V - V0) / a
Plugging in the values, we get:
t = (0 - 11.4) / (-3.80)
t = -11.4 / -3.80
t = 3 seconds
Therefore, the time taken for the car to stop is 3 seconds.