Question

If an object is travelling with initial speed u m/s, accelerating at a rate of a m/s", and

covers a distances m, then its final speed o m/s follows the rule ? = 17 + 2as. Calculate
the distance travelled by a car whose speed increases from 11 m/s to 28 m/s with an acceleration of 3 m/s?.

i need help pls

Answer 1

We are given:

Initial speed, u = 11 m/s
Final speed, v = 28 m/s
Acceleration, a = 3 m/s^2
We need to calculate the distance travelled, s.

We can use the following kinematic equation to solve for the distance travelled, where u is the initial speed, v is the final speed, a is the acceleration, and s is the distance travelled:

v^2 = u^2 + 2as

Substituting the given values, we get:

28^2 = 11^2 + 2(3)s

784 = 121 + 6s

6s = 663

s = 110.5 m

Therefore, the car travels a distance of 110.5 m when its speed increases from 11 m/s to 28 m/s with an acceleration of 3 m/s^2