A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

Answer 1

First, we need to determine the final velocity of the car using the equation:

\[ v = u + at \]

where:
- \( v \) is the final velocity
- \( u \) is the initial velocity (in this case, 0 m/s)
- \( a \) is the acceleration
- \( t \) is the time taken (5.21 seconds)

\[ v = 0 + a \times 5.21 \]
\[ v = 5.21a \]

Next, we can determine the distance traveled by the car using the equation:

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

where:
- \( s \) is the distance traveled (110 m)
- \( u \) is the initial velocity (0 m/s)
- \( a \) is the acceleration
- \( t \) is the time taken (5.21 seconds)

\[ 110 = 0 \times 5.21 + \frac{1}{2} a (5.21)^2 \]
\[ 110 = \frac{1}{2} \times 5.21^2 \times a \]
\[ 110 = 13.52305a \]

Now we can set the two expressions for velocity and distance equal to each other:

\[ 5.21a = 13.52305a \]
\[ 5.21 = 13.52305 \]
\[ a = \frac{5.21}{13.52305} \]
\[ a \approx 0.385 \, \text{m/s}^2 \]

Therefore, the acceleration of the car is approximately \( 0.385 \, \text{m/s}^2 \).