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Papessa [141]
3 years ago
14

Shakina and Juliette set the car's initial velocity to zero and set the acceleration to +1.2 m/s2, then clicked "start." Answer

the following questions.
What is the car's displacement between 0 and 10.0 s?

What was the total distance traveled by the car during this time interval?
Physics
1 answer:
Katarina [22]3 years ago
6 0
Let's start by differentiating the terms distance and displacement. They both refer to the length of paths. Distance only accounts for the total length regardless of the path taken. Displacement measures the linear path from the starting point to the end point. So, it does not necessarily follow the actual path. However, for this problem, assuming that the path is just in one direction, displacement and distance would just be equal. The equation would be:

Distance = Displacement =  v₀t + 0.5at² = 0(10 s) + 0.5(+1.2 m/s²)(10 s)²
Distance = Displacement = 60 meters
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The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

v_x= \frac{dx}{dt}

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dx=v_xdt\\ x=\int v_xdt

Substitute the equation for vₓ =2t² in the integral.

x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m

Substitute 2.0s for t.

x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m

The position of the particle at t =2.0 s is <u>6.4m</u>




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72.1 m

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Hello!

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