A racer starting from rest accelerate in a straight line at a constant rate of 6.25 m/s for 7 seconds. What are the racer velocity at the end of each time? If a parachute is deploy at this time and causes the racer to slow down uniformly at a rate of 2.5 m/s How long will it take the racer to come to stop?
I assume you mean a = 6.25 m/s^2
Please proof read what you type.
v = a t = 6.25 m/s^2 * 7 s
= 43.75 m/s
Now parachute, a = - 2.5 m/s^2 and Vi = 43.75 m/s
v = Vi + a t
0 = 43.75 - 2.5 t
t = 17.5 seconds
To find the velocity of the racer at the end of each time interval, we can use the formula:
v = u + at
Where:
v = final velocity
u = initial velocity (0 in this case, as the racer starts from rest)
a = acceleration (6.25 m/s^2 in this case)
t = time
Let's calculate the velocities at the end of each time interval:
For t = 1 second:
v = 0 + (6.25 m/s) * (1 s) = 6.25 m/s
For t = 2 seconds:
v = 0 + (6.25 m/s) * (2 s) = 12.5 m/s
For t = 3 seconds:
v = 0 + (6.25 m/s) * (3 s) = 18.75 m/s
For t = 4 seconds:
v = 0 + (6.25 m/s) * (4 s) = 25 m/s
For t = 5 seconds:
v = 0 + (6.25 m/s) * (5 s) = 31.25 m/s
For t = 6 seconds:
v = 0 + (6.25 m/s) * (6 s) = 37.5 m/s
For t = 7 seconds:
v = 0 + (6.25 m/s) * (7 s) = 43.75 m/s
Now, let's calculate the time it takes for the racer to come to a stop when the parachute is deployed. We can use the formula:
v = u + at
Where:
v = final velocity (0 m/s in this case, as the racer comes to a stop)
u = initial velocity (43.75 m/s, as this is the velocity at t = 7 s)
a = deceleration (-2.5 m/s^2, as the racer is slowing down uniformly)
t = time
Plugging these values into the formula, we have:
0 = 43.75 m/s + (-2.5 m/s^2) * t
Rearranging the equation:
2.5 t = 43.75 m/s
t = 43.75 m/s / 2.5 m/s^2 = 17.5 s
Therefore, it will take the racer 17.5 seconds to come to a stop after the parachute is deployed.
To find the velocity of the racer at different times during acceleration, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
Initial velocity (u) = 0 (starting from rest)
Acceleration (a) = 6.25 m/s
Time (t) = 7 seconds
To find the velocity at the end of each time, we can substitute the given values into the equation:
v = 0 + (6.25 * 7)
v = 43.75 m/s
Therefore, the racer's velocity at the end of 7 seconds of constant acceleration is 43.75 m/s.
Now, let's find out how long it will take the racer to come to a stop after the parachute is deployed.
Given:
Acceleration due to the parachute (a) = -2.5 m/s (negative because it's deceleration or slowing down)
Initial velocity (u) = 43.75 m/s (velocity at the end of 7 seconds)
Final velocity (v) = 0 (the racer comes to stop)
To find the time required for the racer to come to a stop, we can rearrange the equation:
v = u + at
Rearranging the equation gives us:
t = (v - u) / a
Substituting the given values:
t = (0 - 43.75) / (-2.5)
t = 17.5 seconds
Therefore, it will take the racer 17.5 seconds to come to a stop after the parachute is deployed.