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Vinvika [58]
3 years ago
15

A car traveling east at 40.0 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 25.0 m/s in

3.50 s. (a) What is the magnitude and direction of the car’s acceleration as it slows down? (b) How far does the car travel in the 3.5-s time period?
Mathematics
1 answer:
Anni [7]3 years ago
6 0

Step-by-step explanation:

It is given that,

Initial speed of the car, u = 40 m/s

Final speed of the car, v = 25 m/s

Time taken, t = 3.5 s

(a) We need to find the magnitude and direction of the car’s acceleration as it slows down. It can be calculated using formula as :

a=\dfrac{v-u}{t}

a=\dfrac{25-40}{3.5}

a=-4.28\ m/s^2

The acceleration is in the opposite direction of motion i.e. west.

(b) Let s is the distance the car travel in the 3.5-s time period. It can be calculated using the third equation of motion as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{25^2-40^2}{2\times (-4.28)}

s = 113.9 meters

Hence, this is the required solution.

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What is the average acceleration of a southbound train that slows down from 15 m/s to 8.6 m/s in 1.2 s?
horrorfan [7]

Answer:

The average acceleration of train is 5.34\ m/s^2

Step-by-step explanation:

We have,

A train that slows down from 15 m/s to 8.6 m/s in 1.2 s. It means that 15 m/s is its initial velocity and 8.6 m/s is its final velocity.

It is required to find the average acceleration of the train.

a=\dfrac{v-u}{t}\\\\a=\dfrac{(8.6-15)\ m/s}{1.2\ s}\\\\a=-5.34\ m/s^2

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