A sports car moving at constant velocity travels 110m in 5.1s .If it then brakes and comes to a stop in 4.3s , what is the magnitude of its acceleration (assumed constant) in m/s2?

velociy=110/5.1 m/s

acceleration= (0-110/5.1 )/4.3 m/s^2

-5.01

To find the magnitude of acceleration, we need to find the change in velocity and the time taken for this change.

Given:
Distance traveled before braking = 110m
Time taken to travel this distance = 5.1s

We can find the initial velocity using the formula:

velocity = distance / time

Therefore, initial velocity = 110m / 5.1s = 21.57 m/s

Next, we need to find the final velocity when the car comes to a stop.

Time taken to stop = 4.3s

Using the formula:

final velocity = initial velocity + acceleration * time

Considering the final velocity is zero (as the car comes to a stop), we can rearrange the formula:

acceleration = (final velocity - initial velocity) / time

Substituting the given values:

acceleration = (0 m/s - 21.57 m/s) / 4.3s

Simplifying this, we get:

acceleration = -21.57 m/s / 4.3s

acceleration = -5 m/s^2

The magnitude of the acceleration is the absolute value of the calculated acceleration:

magnitude of acceleration = |-5 m/s^2| = 5 m/s^2

Therefore, the magnitude of the car's acceleration is 5 m/s^2.

To find the magnitude of the acceleration, we can use the kinematic equation:

v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time taken

Since the car is moving at a constant velocity, its acceleration is zero (a = 0) during the first part of the problem, when it travels 110m in 5.1s. So, the initial velocity (u) and final velocity (v) are the same.

Using the equation for distance:

s = ut + (1/2)at^2

We can rewrite it to solve for the initial velocity (u):

u = (2s - at^2) / (2t)

Where:
s = distance traveled
a = acceleration
t = time taken

Substituting the given values into the equation, we get:

u = (2 * 110m - 0 * (5.1s)^2) / (2 * 5.1s)
u = 220m / 10.2s
u = 21.6 m/s

Now, we need to find the acceleration (a) when the car comes to a stop in 4.3 seconds. The initial velocity (u) in this case is 21.6 m/s, and the final velocity (v) is 0 m/s.

Using the kinematic equation:

v = u + at

0m/s = 21.6m/s + a * 4.3s

Rearranging the equation to solve for acceleration (a), we get:

a = (0m/s - 21.6m/s) / 4.3s
a = -21.6m/s / 4.3s
a = -5.02 m/s^2

The magnitude of acceleration is always positive, so the magnitude of the car's acceleration is 5.02 m/s^2.