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Bad White [126]
3 years ago
5

11)A 1100 kg car travels on a straight highway with a speed of 30 m/s. The driver sees a red light ahead and applies her

Physics
1 answer:
cestrela7 [59]3 years ago
3 0

Answer:

Time taken = 8.25 second

Explanation:

Given:

Force = 4000 N

Force = ma

4,000 = (1100)(a)

Acceleration = 3.6363 m/s²

v = u + at

0 = 30 + (3.6363)t

Time taken = 8.25 second

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1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc
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Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of 3.0 m/s^2.

So, the acceleration of the express train, a=-3 m/s^2.

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, a=-3 m/s^2

So, 0=36+(-3)t

\Rightarrow t= 36/3=12 seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

v^2=u^2+2as

\Rightarrow 0^2=36^2+2(-3)s

\Rightarrow s=\frac{36\times36}{6}

\Rightarrow s=216 meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m  [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

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