Given, distance covered by the truck (s) = 400 meters

Time taken to cover the distance (t) = 20 seconds

Initial velocity of the truck (u) = 0 (since it starts from a state of rest)

From the equations of motion, \(s\,=\,ut+\frac{1}{2}at^2\)

\(Therefore,400 = 0(20s) + \frac{1}{2}(a)(400s^2) = 2ms^{-2}\)

The acceleration of the truck is equal to 2ms^{-2}

As per the second law of motion, force = \(Mass\,\times\,Acceleration\)

Mass of the truck = 7 tonnes = 7000 kg

Force acting on the truck = 7000 kg \( \times\) 2 m.s^{-1 }= 14000 kg.m.s^{-2} = 14000 N

Therefore, a force of 14000 N is acting on the truck.

Answered by Shivani Kumari | 2 years agoA motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required

A hammer of mass 500 g, moving at 50 m s^{-1,} strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer

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**The following is the distance-time table of an object in motion:**

Time (seconds) |
Distance (meters) |

0 | 0 |

1 | 1 |

2 | 8 |

3 | 27 |

4 | 84 |

5 | 125 |

6 | 216 |

7 | 343 |

**(a)** What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?

**(b)** What do you infer about the forces acting on the object?

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