Starting from rest a car acquired a velocity of 16m/s in 4s.If mass of the cart is 3kg Find Force acting on it.Also find Distance travelled by the car .
Answers
Answer:
Answer:
Distance covered by car is 32 metres.
Force acting on it is 12 newton.
Explanation:
According to the Question
It is given that
Initial velocity ,u = 0m/s
Final velocity ,v = 16m/s
Time taken ,t = 4s
Mass of cart ,m = 3kg
we need to calculate the force acting on it and distance covered by it .
Firstly we calculate the acceleration of the car .
Using Kinematics Equation
v = u + at
Putting all the value we get
➺ 16 = 0 + a×4
➺ 16 = 4×a
➺ a = 16/4
➺ a = 4m/s²
So, the acceleration of the car is 4m/s².
As we know that Force is calculated by the product of mass and acceleration .
F = ma
substitute the value we get
➺ F = 3×4
➺ F = 12N
Hence, the force acting is 12 newton.
Now, calculating the distance covered by car in given condition.
again using Kinematics Equation
v² = u² + 2as
substitute the value we get
➺ 16² = 0² + 2×4 ×s
➺ 256 = 0 + 8 × s
➺ 256 = 8×s
➺ s = 256/8
➺ s = 32m
Hence, the distance covered by the car is 32 metres.
Step-by-step explanation:
There are two equations provided here.
3x – 5y – 4 = 0 or 3x -5y = 4 ……. (1).
9x = 2y + 7 or 9x -2y = 7 ……(2).
Now only focusing on eq. numbered 1 and 2 …
3x – 5y = 4………(1)
9x – 2y = 7 ……(2)
Multiply eq. (1) by 3, we get….
3 (3x – 5y = 4) or 9x – 15y = 12…… (3).
Now we can easily subtract (2) and (3) to get…..
9x – 2y = 7
– 9x – 15y = 12.
– +13 y = -5
Or we get 13 y = -5 or y = -5/13.
Now we substitute the value of Y = -5/13 in equation (1) we get the value of x.
3x – 5(-5/13) = 4
3x +25/13 = 4
39x + 25 = 52
Or 39x = 52 – 25
39x = 27
Or x = 27/39 = 9 / 13
So x = 9/13 and y = -5/13.