Question

(b) A car starts from rest and travels upwards along a straight road inclined at an angle of
5,0°
+
The length of the road is 450 m and the car has mass 800 kg. The speed of the car
increases at a constant rate and is 28 ms- at the top of the slope.
(i) Determine, for this car travelling up the slope,
1. its acceleration,​

Answer 1

Given:

A car starts from rest and travels upwards along a straight road inclined at an angle of

5°. The length of the road is 450 m and the mass of car is 800 kg. The speed of the car

increases at a constant rate and is 28 ms- at the top of the slope.

To find:

Net acceleration of the car

Calculation:

Let net acceleration of car be a ;

Since acceleration is constant, we can easily applied the equation of kinematics to solve this type of problems.

$\therefore \: {v}^{2} = {u}^{2} + 2as$

$= > {(28)}^{2} = {(0)}^{2} +( 2 \times a \times 450)$

$= > {(28)}^{2} = 2 \times a \times 450$

$= > \: 900a = 784$

$= > a = \dfrac{784}{900}$

$= > \: a = 0.87 \: m {s}^{ - 2}$

So, final answer is:

$\boxed{ \large{ \bold{\: a = 0.87 \: m {s}^{ - 2} }}}$

To Note:

• Net acceleration in this case is different from applied acceleration because net acceleration is obtained by subtracting the weight component [ $g\sin(\theta)$ ] from applied acceleration.