A bus is moving at a speed of 20m/s beyond to slow at a constant rate of 3m/s each second find how far it goes before stopping
To find the distance traveled by the bus before stopping, we need to determine the time it takes for the bus to come to a stop.
We can use the formula:
Final velocity (v) = Initial velocity (u) + (acceleration (a) × time (t))
In this case, the initial velocity (u) is 20 m/s, the acceleration (a) is -3 m/s^2 (as the bus is slowing down), and the final velocity (v) is 0 m/s (as the bus comes to a stop).
0 = 20 + (-3 × t)
Rearranging the equation, we have:
3t = 20
Dividing both sides by 3, we get:
t = 20/3 ≈ 6.67 seconds
Now, we can calculate the distance traveled using the formula:
Distance (d) = (Initial velocity (u) × time (t)) + 0.5 × (acceleration (a) × time (t))^2
Plugging in the values, we have:
d = (20 × 6.67) + 0.5 × (-3 × 6.67)^2
Simplifying further:
d = 133.4 + 0.5 × (-3 × 6.67)^2
d = 133.4 + 0.5 × (-3 × 6.67)^2
d ≈ 133.4 + 0.5 × 100.0469
d ≈ 133.4 + 50.0234
d ≈ 183.43 meters
Therefore, the bus travels approximately 183.43 meters before coming to a stop.
To find out how far the bus goes before stopping, we need to determine the time it takes for the bus to come to a complete stop. We can use the equation of motion:
v = u + at
Where:
v = final velocity (0 m/s when the bus stops)
u = initial velocity (20 m/s)
a = acceleration (in this case, negative 3 m/s^2)
t = time
We rearrange the equation to solve for time:
t = (v - u) / a
Let's substitute the values into the equation and calculate the time it takes for the bus to stop:
t = (0 - 20) / (-3)
t = 20 / 3
t ≈ 6.67 seconds
Now that we know it takes approximately 6.67 seconds for the bus to stop, we can find the distance covered using the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (20 m/s)
t = time (6.67 seconds)
a = acceleration (negative 3 m/s^2)
Plugging in the values:
s = 20 * 6.67 + (1/2) * (-3) * (6.67)^2
s = 133.4 - 66.7
s ≈ 66.7 meters
Therefore, the bus will travel approximately 66.7 meters before coming to a complete stop.
A truck traveling with an initial velocity of 26 m/s begins to slow down. The truck travels for 336 m in the time of 55 s. What is the truck\'s acceleration?
a = -3 m/s^2
v = Vi + a t
at stop
0 = 20 - 3 t
t = 20/3 (they should have started at 21 m/s)
d = Vi t + (1/2) a t^2
d = 20 (20/3) - 3/2 (20^2/3^2)
= 20^2/3 - 20^2/6
= 20^2/6 = 400/6 = 200/3
or easy way for second half is average speed during stop = 10
10 (20/3) = 200/3