Acar is going east at 30 mph along a straight road when the car begins to accelerate at a constant rate. eight seconds later, the car is going east at 65 mph. a) convert each velocity to meters per second. b) find the acceleration of the car. c) how far does the car go during the 8 s interval?

a) 30 mph = 13 m/s. 65 mph = 29 m/s

b) The acceleration of the car is 2 m/s²

c) The car traveled 168 m in the 8 s interval.

Explanation:

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a) Knowing that 1 mile = 1609.34 m and 1 h = 3600 s, let´s convert the velocities to m/s.

30 mi/ 1h · (1609.34 m/ 1 mi) · (1 h / 3600 s) = 13 m/s

65 mi/h  · (1609.34 m/ 1 mi) · (1 h / 3600 s) = 29 m/s

b) The acceleration of the car can be obtained from the equation of velocity:

v = v0 + a · t

Where:

v = velocity of the car at time t

v0 = initial velocity

a = acceleration

t = time

Then, solving for acceleration:

(v - v0)/t = a

a = (29 m/s - 13 m/s) / 8 s = 2 m/s²

c) The position of the car can be calculated using the following equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the car at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

If we place the origin of the frame of reference at the point where the car starts accelerating, then, x0 = 0.

x = 13 m/s · 8 s + 1/2 · 2m/s² · (8 s)²

x = 168 m

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