You take a trip in your car. Starting from rest, you accelerate at 1.00 m/s2 for 10.0 s, heading north. You then stop accelerating and immediately head west at a constant speed for an additional 20.0 s, at which point, you crash into a building coming to an instant stop. At the instant before impact, your average velocity is:
Answer: 6.87 m/s approximately north-west
BUT I am not sure HOW to get down to this answer so please explain !
Thanks !
distance N=1/2 *1*10^2=50m
head west velocity=10*1=10m/s
distance w=10m/s*20 sec=200m
displacement: sqrt(200^2+50^2)
direction approximately NW again
avg velocity(magnitude)=displacement/tim = displacementabove/30sec
To find the average velocity at the instant before impact, we need to break down the motion into its components (north and west) and calculate the average velocity separately for each component.
First, let's calculate the average velocity for the north component of the motion.
1. Starting from rest, you accelerate at 1.00 m/s² for 10.0 s, heading north.
The northward velocity can be calculated using the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since you start from rest, the initial velocity (u) is 0 m/s.
The acceleration (a) is given as 1.00 m/s², and the time (t) is 10.0 s.
Plugging in the values, we have:
v = 0 + (1.00 m/s²)(10.0 s)
v = 10.00 m/s
So, your northward velocity is 10.00 m/s.
Next, let's calculate the average velocity for the west component of the motion.
2. After accelerating for 10.0 s, you stop accelerating and immediately head west at a constant speed for an additional 20.0 s.
Since the acceleration is zero, the velocity remains constant while heading west.
Therefore, the westward velocity remains the same throughout the 20.0 s time interval.
Now, to find the average velocity at the instant before impact, we need to calculate the resultant velocity (v) using the northward and westward velocities.
3. To calculate the resultant velocity, we can use the Pythagorean theorem.
v = √(vn² + vw²),
where vn is the northward velocity and vw is the westward velocity.
Plugging in the values, we have:
v = √((10.00 m/s)² + vw²)
Since we want to find the average velocity right before impact, we know the total time it took for the motion is 10.0 s (accelerating north) + 20.0 s (constant speed west) = 30.0 s.
4. Now, let's calculate the westward displacement (dw) during the 20.0 s heading west.
Since the speed is constant, the displacement can be calculated using:
dw = v × t,
where v is the westward velocity and t is the time interval.
Plugging in the values, we have:
dw = vw × 20.0 s
5. To find the westward velocity (vw), we can rewrite the displacement equation as:
vw = dw / 20.0 s
We can substitute this value of vw back into the resultant velocity (v) equation from step 3.
v = √((10.00 m/s)² + (dw / 20.0 s)²)
6. Now, let's calculate the northward displacement (dn) during the 10.0 s accelerating north.
The formula for displacement with constant acceleration is:
dn = ut + 0.5at²,
where u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values, we have:
dn = (0 m/s)(10.0 s) + 0.5(1.00 m/s²)(10.0 s)²
7. To find the northward velocity (vn), we can rewrite the displacement equation as:
vn = dn / 10.0 s
Substituting this value of vn back into the resultant velocity (v) equation from step 3.
v = √((vn / 10.0 s)² + (dw / 20.0 s)²)
8. Compute the average velocity.
The average velocity is given by the formula:
average velocity = total displacement / total time
The total displacement is the sum of the northward and westward displacements.
total displacement = dn + dw
The total time is the sum of the time taken for acceleration north and the time spent at constant velocity heading west.
total time = 10.0 s + 20.0 s
average velocity = (dn + dw) / (10.0 s + 20.0 s)
Now that we have all the values calculated, we can plug them into the average velocity formula to find the final answer.