A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the driver notices a tree limb that has fallen on the road and brakes hard for t2 = 5 s with a constant acceleration of a2 = -5.93 m/s2.

Write an expression for the car's speed just before the driver begins braking, v1.

If the limb is on the road at a distance of 550 meters from where the car began its initial acceleration, will the car hit the limb?

Well, the expression for the car's speed just before the driver begins braking, v1, can be found 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 the car starts from rest, the initial velocity u is 0. We know the acceleration a1 is 1.76 m/s2 and the time t1 is 20 s. Plugging these values into the equation, we get:

v1 = 0 + 1.76 m/s2 * 20 s
v1 = 35.2 m/s

So, the car's speed just before the driver begins braking is 35.2 m/s.

Now, let's see if the car will hit the limb. The position of the car during the first phase of acceleration can be found using the equation s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Again, u is 0 and a1 is 1.76 m/s2, and we want to find the displacement s1 (distance car traveled during initial acceleration). Plugging in these values and t1 = 20 s, we get:

s1 = 0 + (1/2) * 1.76 m/s2 * (20 s)^2
s1 = 1/2 * 1.76 m/s2 * 400 s^2
s1 = 352 m

So, during the initial acceleration, the car traveled a distance of 352 meters.

Now, we need to calculate the position of the car during the braking phase. In this case, the acceleration a2 is -5.93 m/s2 (negative because it's in the opposite direction) and the time t2 is 5 s. The displacement s2 (distance car traveled during braking) can be found using the equation s = ut + (1/2)at^2. Since the initial velocity u is the final velocity v1 from before, we have:

s2 = 35.2 m/s * 5 s + (1/2) * -5.93 m/s2 * (5 s)^2
s2 = 176 m - 74.25 m
s2 = 101.75 m

So, during the braking phase, the car traveled a distance of 101.75 meters.

Now, we need to add the distances traveled during the initial acceleration and the braking to find the total displacement.

Total displacement = s1 + s2
Total displacement = 352 m + 101.75 m
Total displacement = 453.75 m

Since the distance of the limb is given as 550 meters, and the car's total displacement is less than that, the car will not hit the limb.

No worries, it's safe from the car's path!

To find the expression for the car's speed just before the driver begins braking, v1, we can use the equation:

v1 = v0 + a1 * t1

Since the car starts from rest (v0 = 0) and accelerates with a1 = 1.76 m/s^2 for t1 = 20 s, we can substitute the values into the equation:

v1 = 0 + 1.76 * 20
v1 = 35.2 m/s

Therefore, the car's speed just before the driver begins braking is 35.2 m/s.

To determine if the car will hit the limb on the road, we need to calculate the distance the car covers during the braking phase. The equation to find the distance covered during uniform acceleration is:

d2 = v1 * t2 + 0.5 * a2 * t2^2

Substituting the values, we have:

d2 = 35.2 * 5 + 0.5 * (-5.93) * (5^2)
d2 = 176 - 73.825
d2 = 102.175 m

The distance covered during braking is approximately 102.175 m. Since the limb is at a distance of 550 meters from where the car began its initial acceleration, we can compare these distances:

550 > 102.175

The car will hit the limb on the road since the distance covered during braking is less than the distance to the limb.

To find the expression for the car's speed just before the driver begins braking, v1, we can use the equation of motion:

v1 = u1 + a1 * t1

Here, u1 represents the initial velocity, which is 0 m/s since the car starts from rest. So the equation becomes:

v1 = 0 + a1 * t1
v1 = 0 + 1.76 m/s^2 * 20 s
v1 = 35.2 m/s

Therefore, the car's speed just before the driver begins braking is 35.2 m/s.

To determine if the car will hit the limb at a distance of 550 meters, we need to calculate the distance the car travels during the time it is braking. We can use the equation of motion again:

s = u * t + 0.5 * a * t^2

Since the car is braking, the initial velocity, u2, is the speed just before the driver begins braking, v1. So the equation becomes:

s = v1 * t2 + 0.5 * a2 * t2^2
s = 35.2 m/s * 5 s + 0.5 * (-5.93 m/s^2) * (5 s)^2
s = 176 m - 73.875 m
s = 102.125 m

The car travels a distance of 102.125 meters during the time it is braking. Since this distance is less than the distance to the fallen limb (550 meters), the car will not hit the limb.

V1 = a1*t1 = 1.76 * 20 = 35.2 m/s.

d1 = 0.5a1*t1^2 = 0.88*20^2 =

V2 = V1 + a*t = 35.2 - 5.93*5 = 5.55 m/s. The 5-second deceleration is not enough to bring the car to a full stop.

d2 = V1*t + 0.5a*t^2.
d2 = 35.2*5 - 0.5*5.93*5^2 =

d = d1+d2 =