Question

a truck of mass 4 x 10 power 7 kg is pulled by a force of 6 x 10 power 5 Newton calculate the acceleration find its velocity after 10 minutes if initially the truck was at rest​

Answer 1

Answer:

The required acceleration is 0.015 m/s².

The required velocity is 9 m/s.

Step-by-step explanation:

Mass, m = 4 × 10⁷ kg

Force, F = 6 × 10⁵ N

From Newton's second law,

F = ma

Substituting the values,

$\sf 6 \times 10^5 = 4 \times 10^7 \times a \\\\ \sf a = \dfrac{6 \times 10^5}{4 \times 10^7} \\\\ \sf a = 1.5 \times 10^{-2} \\\\ \sf a = 0.015 m/s^2$

The acceleration of the truck is 0.015 m/s².

______________________

Initially, the truck was at rest.

Initial velocity, u = 0 m/s

time, t = 10 min = 10 × 60 s = 600 s

Final velocity, v = ?

First equation of motion :

v = u + at

Substituting,

v = 0 + 0.015(600)

v = 9 m/s

The velocity of the truck after 10 min is 9 m/s.

Answer 2

$\orange{\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given : \\ \\ \hookrightarrow \: \sf \:mass \: of \: the \: truck , \: \: m = 4 \times {10}^{7} \\ \\\hookrightarrow \: \sf \:force , \: \: F = 6 \times {10}^{5} \: N\\ \\ \hookrightarrow \: \sf \:initial \: velocity, \: \: u = 0 \: \: \: [ \bf \: at \: rest] \\ \\ \hookrightarrow \: \sf \:time, \: t = 10 \: min \\ = 10 \times 60 \: s \\ = 600 \: s\\ \\ \\ \blue{ \underline{ \pink{\bf \: we \: have \: to \: find}}} \\ \\ \leadsto \: \sf \: acceleration \: of \: the \: truck = a \\ \\ \small{ \leadsto \: \sf \: final \: velocity \: of \: the \: truck \: after \: 10 \: min = v}\end{array}}}}$

${\boxed{\boxed{\begin{array}{cc} \red{ \underline{\bf \: solution}} \\ \\ \text{we \: know \: that} , \\ \\ \: \: \: \: \: \: \: \: \: \: \sf \: F = ma \\ \\ \sf \: \implies \:a = \frac{F}{m} \\ \\ = \frac{6 \times {10}^{5} }{4 \times {10}^{ 7} } \\ \\ = \frac{3}{2} \times {10}^{5- 7} \\ \\ = 1.5 \times {10}^{ - 2} \\ \\ = 0.015 \: \sf \: m {s}^{ - 2} \\ \\ \orange{ \boxed{\therefore \: \sf \: acceleration, \: a = 0.015 \: m {s}^{ - 2} }}\end{array}}}}$

Again,

${\boxed{\boxed{\begin{array}{cc}\bf \: we \: know \: that: \\ \\ \sf \: v = u + at \\ \\ = 0 + 0.015 \times 600 \\ \\ = \frac{1.5}{100} \times 600 \\ \\ = 1.5 \times 6 \\ \\ = 9 \: ms {}^{ - 1} \\ \\ \orange{ \boxed{ \therefore \sf \: velocity, \: \: v = 9 \: m {s}^{ - 1}}} \end{array}}}}$