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velikii [3]
3 years ago
9

A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .

Physics
1 answer:
Tju [1.3M]3 years ago
3 0

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

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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
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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

Learn more about accelerated motion:

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