A car accelerates uniformly from rest, travelling 82.5 m in the first 6.4 s. Calculate its acceleration in ms 2 to 3d.p.

To find the acceleration of the car, we can use one of the kinematic equations for uniformly accelerated motion. Since the car starts from rest, the initial velocity \( u = 0 \). The displacement \( s = 82.5 \, \text{m} \), the time \( t = 6.4 \, \text{s} \), and we are asked to find the acceleration \( a \).

The appropriate kinematic equation is:

\[
s = ut + \frac{1}{2}at^2
\]

Substitute the known values into the equation:

\[
82.5 = 0 \times 6.4 + \frac{1}{2}a(6.4)^2
\]

Simplifying:

\[
82.5 = \frac{1}{2}a(40.96)
\]

Now solve for \( a \):

\[
82.5 = 20.48a
\]

\[
a = \frac{82.5}{20.48}
\]

\[
a \approx 4.027 \, \text{m/s}^2
\]

Thus, the acceleration of the car is approximately \( 4.027 \, \text{m/s}^2 \) to 3 decimal places.

Would you like more details on this calculation or any other aspect?

Here are 5 related questions to consider:
1. What would be the velocity of the car after 6.4 seconds?
2. How long would it take for the car to travel 150 meters at the same acceleration?
3. What distance would the car travel if it accelerated for 10 seconds?
4. How would the result change if the car started with an initial velocity?
5. How is acceleration related to the force applied to the car?

**Tip**: For uniformly accelerated motion, always check if the initial velocity is zero to simplify calculations!

A car accelerates uniformly from rest, travelling 82.5 m in the first 6.4 s. Calculate its acceleration in ms^2 to 3 decimal places.

Learn how to calculate the acceleration of a car that accelerates uniformly from rest, covering 82.5 meters in 6.4 seconds. This detailed step-by-step guide uses kinematic equations to find the acceleration to 3 decimal places. Ideal for students in Grades 9-12 studying motion and algebra.

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