Explanation:
Given data
velocity v= 25m/s
The time it takes to put on brake t= 0.3s
the distance covered when the brake was put on is
v=s/t
s= v*t
s= 25*0.3s
s= 7.5m
hence the distance covered is 7.5m
Also the rate of decrease in aceleration is 5m/s^2
we can also calculate the distance covered at this rate
v^2=u^2+2as
25^2= 0+2*5*s
625=10s
divide both sides by 10
s=625/10
s= 62.5m
The total distance covered between putting on the brakes and decelareation is 7.5+62.5= 70m
Given that the tree is 75m ahead, the car would not hit the tree
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Answer:
F = - 3.53 10⁵ N
Explanation:
This problem must be solved using the relationship between momentum and the amount of movement.
I = F t = Δp
To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio
v = d / t
t = d / v
Reduce SI system
m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg
d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m
Let's calculate
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
With this value we use the momentum and momentum relationship
F t = m v - m v₀
As the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵
F = - 3.53 10⁵ N
The negative sign indicates that the force is exerted against the bullet
Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.