Question

Dynamics of Rigid Bodies:

Problem 1) A car is moving with constant acceleration covers 450 m in a 5 second interval and further it covers 700 m in a 10 second interval. What is the acceleration of the car?

Given:
D/S1= 450 m Total Distance = 450+700 = 1150
T1= 5 seconds
Distance/S2=700m
T2= 10 seconds

when car covers 450m
Formula: ( S=ut + 1/2 at^2)
450m = 5u + 1/2 x a x (5)^2
450m = 5(-5a+1150)+1/2 a (25)
450m = -25+ 575 + 25/2
-575+450=-25/2 a
-125 = -12.5 a
a = +10 m/s^2

Get the value of u :
1150 = 10u + 1/2 x a x (10)^2
1150 = 10u + 1/2 x a x 100
1150 = 10u + 50a
-10u= 50 a - 1150
10u/50a= -1150
u = -5 +1150

Answer 1

The acceleration of the car is +10 m/s^2.

Step-by-step explanation:

Using the formula S = ut + 1/2at^2, we can calculate the acceleration of the car.

When the car covers 450 m in 5 seconds, we have:

450 = 5u + 1/2 x a x 5^2

Simplifying this equation gives us:

450 = 5u + 12.5a

Next, when the car covers a total distance of 1150 m in 15 seconds, we have:

1150 = 10u + 1/2 x a x 10^2

Simplifying this equation gives us:

1150 = 10u + 50a

We can now solve for u in terms of a using the first equation:

5u = 450 - 12.5a

u = (450 - 12.5a)/5

Substituting this expression for u into the second equation gives:

1150 = 2(450 - 12.5a) + 50a

Simplifying and solving for a gives:

a = 10 m/s^2

Therefore, the acceleration of the car is +10 m/s^2.