A car travelling at 90 km/h is brought to rest over a distance of 120 m. What is the acceleration in metres per second squared?

A car travelling at 90 km/h is brought to rest over a distance of 120 m. What is the acceleration in metres per second squared?

• A. 2.6
• B. -2.6
• C. 1.3
• D. -1.3

Answer: (B)

To find the acceleration when a car is brought to rest, we can use the formula that relates acceleration, initial velocity, final velocity, and distance. The formula is: \[ v^2 = u^2 + 2as \] Where: - \( v \) is the final velocity (0 m/s since the car comes to rest) - \( u \) is the initial velocity (90 km/h, which is converted to meters per second by using the conversion factor of 1 km/h = 1/3.6 m/s, resulting in 25 m/s) - \( a \) is the acceleration - \( s \) is the distance (120 m) Starting with the information, we can rearrange the equation to solve for acceleration \( a \): \[ 0 = (25)^2 + 2a(120) \] This simplifies to: \[ 0 = 625 + 240a \] From here, we can isolate \( a \): \[ 240a = -625 \] \[ a = -\frac{625}{240} \] Calculating that gives \( a \approx -2.6 \) m/s². The negative sign indicates that the acceleration is

To find the acceleration when a car is brought to rest, we can use the formula that relates acceleration, initial velocity, final velocity, and distance. The formula is:

[ v^2 = u^2 + 2as ]

Where:

• ( v ) is the final velocity (0 m/s since the car comes to rest)

• ( u ) is the initial velocity (90 km/h, which is converted to meters per second by using the conversion factor of 1 km/h = 1/3.6 m/s, resulting in 25 m/s)

• ( a ) is the acceleration

• ( s ) is the distance (120 m)

Starting with the information, we can rearrange the equation to solve for acceleration ( a ):

[ 0 = (25)^2 + 2a(120) ]

This simplifies to:

[ 0 = 625 + 240a ]

From here, we can isolate ( a ):

[ 240a = -625 ]

[ a = -\frac{625}{240} ]

Calculating that gives ( a \approx -2.6 ) m/s².

The negative sign indicates that the acceleration is