A motorist is speeding along at 120 km/hr when he sees a squirrel on the road 200 meters in front of him. He tries to stop, but it takes 12 seconds for his car to stop.
(a) What is the acceleration of the car? (assume acceleration was constant)
(b) Does the squirrel survive?
(c) How fast was the car moving at 100 meters?

A.) v = v0 + at Use t = 12 seconds and v0 = 120 km/hr. First, notice the stopping time is in seconds, but the velocity is per hour. We will also need the distance to be in meters, so convert the velocity to m/s: v0= 120 km/1 hr * 1000m/1km* 1hr/3600s= 33.33/1s v0 = 33.33 m/s The car is stopped at the end, so the final velocity is equal to zero. 0 = 33.33 m/s + a(12 s) -33.33 m/s = a(12 s) a = -2.78 m/s2 Note the acceleration is negative. This means it is slowing down the vehicle as motion progresses in the positive direction. Just what you would expect in a problem where the vehicle is

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Question|Asked by Taurus28

Asked 271 days ago|2/9/2024 12:10:26 PM

Updated 271 days ago|2/9/2024 12:42:06 PM

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A.) v = v0 + at

Use t = 12 seconds and v0 = 120 km/hr. First, notice the stopping time is in seconds, but the velocity is per hour. We will also need the distance to be in meters, so convert the velocity to m/s:

v0= 120 km/1 hr * 1000m/1km* 1hr/3600s= 33.33/1s

v0 = 33.33 m/s

The car is stopped at the end, so the final velocity is equal to zero.

0 = 33.33 m/s + a(12 s)

-33.33 m/s = a(12 s)

a = -2.78 m/s2

Note the acceleration is negative. This means it is slowing down the vehicle as motion progresses in the positive direction. Just what you would expect in a problem where the vehicle is slowing down.

Use t = 12 seconds and v0 = 120 km/hr. First, notice the stopping time is in seconds, but the velocity is per hour. We will also need the distance to be in meters, so convert the velocity to m/s:

v0= 120 km/1 hr * 1000m/1km* 1hr/3600s= 33.33/1s

v0 = 33.33 m/s

The car is stopped at the end, so the final velocity is equal to zero.

0 = 33.33 m/s + a(12 s)

-33.33 m/s = a(12 s)

a = -2.78 m/s2

Note the acceleration is negative. This means it is slowing down the vehicle as motion progresses in the positive direction. Just what you would expect in a problem where the vehicle is slowing down.

Added 271 days ago|2/9/2024 12:40:17 PM

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B.) In order to find out if the squirrel survives, we need to know how far the vehicle travelled before it stops. If the distance travelled is less than the distance to the squirrel, the squirrel will survive. Use equation 1 from above for this part.

x = x0 + v0t + ½at2

Plug in the acceleration from part a) and the initial conditions.

x = 0 m + (33.33 m/s)(12 s) + ½(-2.78 m/s2)(12 s)2

x = 399.96 m – 200.16 m

x = 199.8 m

The distance the vehicle took to stop was less than 200 m, so the squirrel did survive the encounter…barely.

x = x0 + v0t + ½at2

Plug in the acceleration from part a) and the initial conditions.

x = 0 m + (33.33 m/s)(12 s) + ½(-2.78 m/s2)(12 s)2

x = 399.96 m – 200.16 m

x = 199.8 m

The distance the vehicle took to stop was less than 200 m, so the squirrel did survive the encounter…barely.

Added 271 days ago|2/9/2024 12:40:36 PM

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C.) v^2 = v0^2 + 2a(x – x0)

Use x = 100 meters and the units of the velocities in m/s and acceleration in m/s2.

v2 = (33.33 m/s)^2 + 2(-2.78 m/s2)(100 m – 0 m)

v2 = 1110.89 m^2/s^2 – 556 m^2/s^2

v2 = 554.89 m^2/s^2

v = 23.56 m/s

The vehicle was moving at 23.56 m/s (84.8 km/hr) at the 100 meter mark.

Use x = 100 meters and the units of the velocities in m/s and acceleration in m/s2.

v2 = (33.33 m/s)^2 + 2(-2.78 m/s2)(100 m – 0 m)

v2 = 1110.89 m^2/s^2 – 556 m^2/s^2

v2 = 554.89 m^2/s^2

v = 23.56 m/s

The vehicle was moving at 23.56 m/s (84.8 km/hr) at the 100 meter mark.

Added 271 days ago|2/9/2024 12:42:06 PM

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Added 271 days ago|2/9/2024 12:41:24 PM

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