Question:medium

An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?

Updated On: Jan 19, 2026
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Solution and Explanation

Given: 

Mass of vehicle, \( m = 1500 \, \text{kg} \)
Negative acceleration (deceleration), \( a = 1.7 \, \text{m/s²} \)

Concept:

According to Newton’s second law, the force required to produce an acceleration is:
\[ F = m \cdot a \]

Step 1: Calculation

\[ F = 1500 \times 1.7 \]

\[ F = 2550 \, \text{N} \]

Answer:

The force required between the vehicle and the road to stop it = 2550 N (opposite to the direction of motion)

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Questions Asked in CBSE Class IX exam

(Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.

All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find: 

(i) how many cross - streets can be referred to as (4, 3). 

(ii) how many cross - streets can be referred to as (3, 4).

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