Speedy Sue is driving at 50 m/s. The roadway is covered with ice when she sees d

Question

Speedy Sue is driving at 50 m/s. The roadway is covered with ice when she sees directly in front of her, a van, 1 km away traveling at 10 m/s. She slams on her brakes, but the anti-lock braking system kicks in, and she can only slow with an acceleration of - 0.1 m/s.

Does she hit the van?

A. Yes

B. No

C. Cannot tell from this information

D. Yes, if the ice doesn't melt

If she hits the van, when does the collision occur?

A. 774 s after she sees the van

B. 25.8 s after she sees the van

C. 225 s. after she sees the van

D. She stops before hitting the van

If she hits the van, what distance from her starting point does she hit it?

A. 258 m

B. 1258 m

C. 5216 m

D. She stops before hitting the van

If she hits the van, how far has the van traveled since she saw it?

A. 258 m

B. 1258 m

C. 5216 m

D. She stops before hitting the van

If she misses the van, but continues to brake, when will the van pass her?

A. 774 s after she passes the van

B. 748 s after she passes the van

C. 225 s after she passes the van

D. She stops before passing the van

Explanation / Answer

For Sue,

Initial velocity of car = vis= 50m/s

Initial position = Xis = 0 m

Acceleration = as = -0.1 m/s^2

If the collision occurs final position = Xf

Kinematic equation for Sue,

Xf – Xis = vis*t + ½*as*t^2

Xf – 0 = 50t-1/2*0.1*t^2

Xf = 50t-1/2*0.1*t^2-------------------(1)

For van,

Initial velocity of car = viv= 10m/s

Initial position = Xiv = 1000 m

Acceleration = av = 0.00 m/s^2

If the collision occurs final position = Xf

Kinematic equation for Sue,

Xf – Xiv = viv*t + ½*av*t^2

Xf – 1000 = 10t

Xf = 1000+10t-------------------(2)

Plug (2) in (1),

1000+10t = 50t-1/2*0.1*t^2

t = 25.8s or t= 774.16s

collision is always occur earlier time hence they will collide at 25.s

Yes Sue hits the van.

Collision occurs at 25.8 s after she sees the van

Position of sue when the collision occur = 50t-1/2*0.1*t^2

Put t=25.8s

Xf = 50t-1/2*0.1*t^2

Xf = 50*25.8 – ½*0.1*25.8^2 = 1258 m            

Distance of van traveled = Xf – 155 = 1258 – 100 = 258 m

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