Question

A tractor, starting from a stationary position, drives down the road. It accelerates at 4 m/s2 for 10 seconds, then travels at a steady pace for another 30 seconds. How much distance has it covered since it started driving?​

A. 1,550 m
B. 1,400 m
C. 200 m
D. 1,200 m

Answer 1

To find the total distance covered by the tractor, both the distance during acceleration (200 m) and at a constant velocity (1200 m) are calculated and added together, resulting in a total distance of 1400 m, which corresponds to option B.

Step-by-step explanation:

Calculating Total Distance Covered by the Tractor

To calculate the total distance covered by the tractor, we will divide the problem into two parts: the distance covered during acceleration and the distance covered while moving at a constant velocity.

Distance During Acceleration

The formula for distance covered during acceleration (starting from rest) is given by d = \(\frac{1}{2}at^2\), where a is the acceleration and t is the time.

Given a = 4 m/s2 and t = 10 s, we can calculate the distance as:

d = \(\frac{1}{2} \times 4 \text{m/s}^2 \times (10 \text{s})^2\) = \(\frac{1}{2} \times 4 \times 100\) = 200 m.

Distance at Constant Velocity

The velocity of the tractor after acceleration is v = at = 4 m/s2 \(\times 10\text{s}) = 40 m/s. Now, the tractor moves at this velocity for 30 seconds.

The distance covered at constant velocity is given by d = vt, where v is the velocity and t is the time = 40 m/s \(\times 30 \text{s}) = 1200 m.

Total Distance Covered

To find the total distance covered, we add the two distances: 200 m + 1200 m = 1400 m. Therefore, the correct answer is B. 1,400 m.

Answer 2

B. 1400 meters.

Step-by-step explanation:

According to the statement, there are two stages for the motion of the tractor, in which we need to find the travelled distance:

(i) Uniform accelerated motion.

`s_(i) = s_(o,i) +v_(o,i)\cdot t_(i) + (1)/(2)\cdot a\cdot t_(i)^(2)` (1)

`v_(i) = v_(o,i)+a\cdot t_(i)` (2)

Where:

`s_(o,i)` - Initial position of the tractor, measured in meters.

`s_(i)` - Final position of the tractor, measured in meters.

`v_(o,i)` - Initial velocity of the tractor, measured in meters per second.

`t_(i)` - Time, measured in seconds.

`a` - Acceleration, measured in meters per square second.

`v_(i)` - Final velocity of the tractor, measured in meters per second.

(ii) Uniform motion.

`s_(ii) = s_(i) + v_(i)\cdot t_(ii)` (3)

Where:

`s_(ii)` - Final position of the tractor, measured in meters.

`t_(ii)` - Time, measured in seconds.

The distance covered by the tractor (
`\Delta s`), measured in meters, is:

`\Delta s = s_(ii)` (4)

If we know that
`a = 4\,(m)/(s^(2))`,
`t_(i) = 10\,s`,
`v_(o,i) = 0\,(m)/(s)`,
`s_(o,i) = 0\,m` and
`t_(ii) = 30\,s`, then distance covered by the tractor is:

By (1) and (2):

`s_(i) = 0\,m + \left(0\,(m)/(s) \right)\cdot (10\,s)+ (1)/(2)\cdot \left(4\,(m)/(s^(2)) \right) \cdot (10\,s)^(2)`

`s_(i) = 200\,m`

`v_(i) = \left(0\,(m)/(s) \right)+\left(4\,(m)/(s^(2)) \right)\cdot (10\,s)`

`v_(i) = 40\,(m)/(s)`

By (3):

`s_(ii) = 200\,m + \left(40\,(m)/(s) \right)\cdot (30\,s)`

`s_(ii) = 1400\,m`

By (4):

`\Delta s = 1400\,m`

The correct answer is B.