The speed of train is reduced uniformly from 15 m/s to 7 m/s while traveling a d

Question

The speed of train is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m. Compute the acceleration. How much farther will the train travel before coming to rest, provided the acceleration remains constant. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.60 m/s^2 for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree? Two cars momentarily stopping at red light signal. At green light, both cars start accelerating from

Explanation / Answer

here,

initial speed , u = 15 m/s

final speed , v = 7 m/s

distance , s= 90 m

a)

let the accelration be a

using third equation of motion

v^2- u^2 =2 * a *s

7^2 - 15^2 = 2 * a * 90

a = - 0.97 m/s^2

b)

let distance travelled before coming to rest be d

using third equation of motion

v^2- u^2 = 2 * a *d

0 - 7^2 = - 2 * 0.97 * d

d = 25.1 m

6)

accelration , a = - 5.6 m/s^2

time , t = 4.2 s

s = 62.4 m

s = u * t + 0.5 * a * t^2

62.4 = u * 4.2 - 0.5 * 5.6 * 4.2^2

u = 26.62 m/s

final speed , v = u + a * t

v = 26.62 - 5.6 * 4.2

v = 3.1 m/s

the final speed of ca before hitting the tree is 3.1 m/s

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