Question

1. a lorry moves with a constant speed of 25 m/s for 12 s, then decelerates uniformly to rest in 5 s. (a) calculate the deceleration. (b) how far does the lorry travel while decelerating? (c) find the total distance travelled. 5. a car accelerates uniformly from 10 m/s to 30 m/s in 4 s. (a) find the acceleration. (b) calculate the distance travelled in this time.

Explanation:

Step1: Calculate the deceleration for the lorry in question 4(a)

Use the formula $a=\frac{v - u}{t}$, where $v = 0$ (final - velocity, comes to rest), $u = 25$ m/s (initial - velocity), and $t = 5$ s (time of deceleration). So, $a=\frac{0 - 25}{5}=- 5$ m/s².

Step2: Calculate the distance travelled while decelerating for the lorry in question 4(b)

Use the formula $s=ut+\frac{1}{2}at^{2}$. Here, $u = 25$ m/s, $a=-5$ m/s², and $t = 5$ s. So, $s=25\times5+\frac{1}{2}\times(-5)\times5^{2}=125 - 62.5 = 62.5$ m.

Step3: Calculate the distance travelled during the constant - speed phase for the lorry in question 4

During the constant - speed phase, $s_1=v\times t$, where $v = 25$ m/s and $t = 12$ s. So, $s_1=25\times12 = 300$ m.

Step4: Calculate the total distance travelled for the lorry in question 4(c)

The total distance $s_{total}=s_1 + s$, where $s_1 = 300$ m and $s = 62.5$ m. So, $s_{total}=300 + 62.5=362.5$ m.

Step5: Calculate the acceleration for the car in question 5(a)

Use the formula $a=\frac{v - u}{t}$, where $v = 30$ m/s, $u = 10$ m/s, and $t = 4$ s. So, $a=\frac{30 - 10}{4}=5$ m/s².

Step6: Calculate the distance travelled for the car in question 5(b)

Use the formula $s=ut+\frac{1}{2}at^{2}$, where $u = 10$ m/s, $a = 5$ m/s², and $t = 4$ s. So, $s=10\times4+\frac{1}{2}\times5\times4^{2}=40 + 40=80$ m.

Answer:

4(a): - 5 m/s²
4(b): 62.5 m
4(c): 362.5 m
5(a): 5 m/s²
5(b): 80 m