A driver of a car that is going 25 m/s suddenly notices a train blocking the road. At the instant the brakes are applied, the train is 60 m away. The car decelerates uniformily and stikes the train 3 secs later. (a) How fast was the car moving on impact? (b) What was the magnitude of its acceleration during the 3 secs?

d = Vo*t + 0.5at^2 = 60m/s.

25*3 + 0.5a(3)^2 = 60,
75 + 4.5a = 60,
4.5a = -15,
a = -3.3333m/s^2.

Vf = Vo + at = 25 + (-3.3333)3 = 15m/s.

To find the answers to the given questions, we can make use of the equations of motion. First, let's define the variables:

v0 = initial velocity of the car (25 m/s)
d = initial distance between the car and the train (60 m)
t = time taken before impact (3 seconds)
vf = final velocity of the car (to be found)
a = acceleration of the car (to be found)

(a) How fast was the car moving on impact?

We need to find the final velocity of the car (vf) when it strikes the train.

We can use the equation of motion:

vf = v0 + at

We know that the acceleration (a) is constant as the car decelerates uniformly. Since the car is slowing down, the acceleration will be in the opposite direction to the initial velocity. Therefore, the acceleration will be negative.

Plugging in the given values:

vf = 25 m/s + a × 3 s

Now, we can use another equation to relate the final velocity (vf), initial velocity (v0), acceleration (a), and displacement (d):

vf^2 = v0^2 + 2ad

Plugging in the given values:

(vf)^2 = (25 m/s)^2 + 2 × a × (-60 m) [Note: Distance (d) is taken negative as it's in the opposite direction to the car's initial velocity]

Simplifying:

(vf)^2 = 625 m^2/s^2 - 120a m^2/s^2

Substituting the value of vf from the first equation:

(25 m/s + a × 3 s)^2 = 625 m^2/s^2 - 120a m^2/s^2

Expanding the equation:

625 m^2/s^2 + 150a m/s + a^2 × (3 s)^2 = 625 m^2/s^2 - 120a m^2/s^2

Simplifying and canceling out terms:

9a^2 - 270a = 0

Factoring out 'a':

a(9a - 270) = 0

Now, since the car is decelerating, the acceleration will be negative, so 'a' cannot be zero. Therefore:

9a - 270 = 0
9a = 270
a = 270/9
a = 30 m/s^2 (acceleration of the car)

Substituting this value of 'a' into the first equation:

vf = 25 m/s + 30 m/s^2 × 3 s
vf = 115 m/s

Therefore, the car was moving at a speed of 115 m/s on impact with the train.

(b) What was the magnitude of its acceleration during the 3 secs?

We have already found the acceleration of the car, which is -30 m/s^2 (negative due to deceleration).

So, the magnitude of acceleration during the 3 seconds is 30 m/s^2.