Answer:

Explanation:
Given that,
Initial speed of a car, u = 13 m/s
Final speed of a car, v = 25 m/s
Time, t = 5 s
We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :

So, the acceleration of the car is
.
Answer:
the focal length of the mirror is : 
Explanation:
Use the formula for the formation of image using a divergent mirror and recalling that the image (s') that this mirror formed is virtual, so it is entered as a negative number in the formula. Use the object position (s) as 10, the image position (s') as -2, and derive the value of the focal length:

2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s

a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)


2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
closed and unsecure circuit