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iren2701
16 days ago
12

A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera

tion (magnitude only) of the jogger, (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?
Physics
1 answer:
Sav [1.1K]16 days ago
4 0

Answer:

a) The jogger's acceleration is 1.5 m/s²

b) The car's acceleration is also 1.5 m/s²

c) Yes, the car covers a distance 76 m greater than the jogger.

Explanation:

a) Acceleration is the change in velocity over a given time interval:

a = (final velocity - initial velocity) / time

For the jogger:

a = (3.0 m/s - 0 m/s) / 2.0 s = 1.5 m/s²

b) For the car:

a = (41.0 m/s - 38.0 m/s) / 2.0 s = 1.5 m/s²

c) To find how far the car has traveled after 2 seconds, use the formula for position under acceleration along a straight path:

x = x₀ + v₀ t + ½ a t²

where

x = position at time t

x₀ = initial position

v₀ = initial velocity

t = elapsed time

a = acceleration

Assuming x₀ = 0 (origin at car's starting point):

x = 38.0 m/s × 2 s + ½ × 1.5 m/s² × (2.0 s)²

x = 79 m

Similarly, position of the jogger after 2 seconds is:

x = 0 m/s × 2 s + ½ × 1.5 m/s² × (2.0 s)² = 3 m

The difference traveled by the car compared to the jogger is 79 m - 3 m = 76 m

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