Question

A toy car starts from rest 0.650m from the edge of a table. The car accelerates at 2.50 m/s^2 until it leaves the table. The toy car falls 1.10 m to the floor. a. How fast is the toy car going when it reaches the table edge? (ans: 3.25 m/s) b. Solve for the range of the toy car after it leaves the table. (ans: 1.54 m)

Answer 1

Answer:

a) The velocity of the toy car is 1.80 m/s when it reaches the edge of the table.

b) The range of the toy car is 0.853 m.

Explanation:

a) While the car is traveling in a straight line with constant acceleration the equation of the velocity and position will be:

v = v0 + a · t

x = x0 + v0 · t + 1/2 · a · t²

Where:

v = velocity at time "t".

v0 = initial velocity.

a = acceleration.

t = time.

x = position at time "t".

x0 = initial positon.

First, let´s find the time it takes the car to travel the distance of 0.650 m on the table (placing the origin of the frame of reference at the point where the car starts, x0 = 0):

x = x0 + v0 · t + 1/2 · a · t²

0.650 m = 0 m + 0 m/s · t + 1/2 · 2.50 m/s² · t²

2 · 0.650 m / 2.50 m/s² = t²

t = 0.721 s

Now, using the equation of the velocity, we can calculate the velocity when the car reaches the table edge:

v = v0 + a · t

v = 0 m/s + 2.50 m/s² · 0.721 s

v = 1.80 m/s

The velocity of the toy car is 1.80 m/s when it reaches the edge of the table.

b) Please, see the attached figure for a better understanding of the problem. The origin of the frame of reference is located at the edge of the table.

The position of the car while it is falling can be calculated using this equation:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the car reaches the ground, the y-component of the position vector is -1.10 m (see "ry final" in the figure). Using that information, we can calculate the time it takes the car to reach the ground. With that time, we can calculate then the range:

y = y0 + v0y · t + 1/2 · g · t²

-1.10 m = 0 + 0 - 1/2 · 9.8 m/s² · t²        (why is v0y = 0? hint: see the direction of the vector velocity when the car reaches the edge of the table. In red in the figure).

-1.10 m / -1/2 · 9.8 m/s² = t²

t = 0.474 s

Now, using the equation of the x-component of the vector r, we can obtain the range:

x = x0 + v0x · t

x = 0 + 1.80 m/s · 0.474 s

x = 0.853 m

The range of the toy car is 0.853 m.