Question

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 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 until it comes to a stop. How far, in meters, from the original location of the limb will the car be when it stops?

Answer 1

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$

where:

$v_1=32.4 m/s$ is the initial velocity at the beginning of the 2nd phase

$v_2=0$ is the final velocity (the car comes to a stop)

$t_2=5 s$ is the time elapsed in the 2nd phase

Substituting,

$s_2=\frac{32.4+0}{2}(5)=81 m$

So, the total displacement of the car is

$s=s_1+s_2=324+81=405 m$

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